不等式

均值不等式

\frac{n}{\sum\limits_{i = 1}^n \frac{1}{x_i} } \le \sqrt[n]{\prod\limits_{i = 1}^n x_i} \le \frac{\sum\limits_{i = 1}^n x_i }{n} \le \sqrt {\frac{\sum\limits_{i = 1}^n {x_i}^2 }{n}}\\\sqrt {x\cdot y}\leq \frac{x-y}{\ln x-\ln y}\leq \frac {x+y}{2}

柯西不等式

\sum\limits_{i = 1}^n {a_i}^2 \cdot \sum\limits_{i = 1}^n {b_i}^2 \ge {\left( {\sum\limits_{i = 1}^n {a_i} \cdot {b_i} } \right)^2}\\ \left( \frac {a_{1}}{b_{1}}={\frac {a_{2}}{b_{2}}=\cdots =\frac {a_{n}}{b_n}} \right) \\f\left( {\frac{{a + b}}{2}} \right) \leqslant \frac{1}{{b - a}}\int_a^b f (x){\text{d}}x \leqslant \frac{f(a) + f(b)}{2}

对数均值不等式

\sqrt {x\cdot y}\leq \frac{x-y}{\ln x-\ln y}\leq \frac {x+y}{2}

琴生不等式

{\textstyle \sum_{i}^np_i=1},f''(x)\geqslant0 \\\sum\limits_{i = 1}^n {{p_i}} f\left( {{x_i}} \right) \geqslant f\left( {\sum\limits_{i = 1}^n {{p_i}} {x_i}} \right) \\\text{Proof:}\;记A = \sum\limits_{i = 1}^n {{p_i}} {x_i}\\ \sum\limits_{i = 1}^n {{p_i}} f\left( {x_i} \right) - f\left( A \right)\\ \begin{align} &= \sum\limits_{i = 1}^n {\left[ {p_if\left( {{x_i}} \right) - {p_i}f\left( A \right)} \right]} \\ &= \sum\limits_{i = 1}^n {p_i}\int\limits_A^{x_i} {f'(x)} {\text{d}}x\\ & \geqslant \sum\limits_{i = 1}^n p_if'(A)({x_i} - A)=0\end{align}

f\left( {\frac{{a + b}}{2}} \right) \leqslant \frac{1}{{b - a}}\int_a^b f (x){\text{d}}x \leqslant \frac{{f(a) + f(b)}}{2}

\begin{gathered} a\ln a + b\ln b \geqslant (a + b)\ln \frac{a + b}{2} \hfill \\ {x_1}^{p_1} \cdot \cdot \cdot {x_n}^{p_n} \leqslant {p_1}{x_1} + \cdot \cdot \cdot + {p_n}{x_n}\left( {\sum\limits_{i = 1}^n {p_i} = 1} \right) \hfill \\ a,b > 0,p > 1,\frac{1}{p} + \frac{1}{q} = 1,ab \leqslant \frac{1}{p}{a^p} + \frac{1}{q}{a^q} \hfill \\ x \geqslant - 1,a > 1/a < 0,{(1 + x)^a} \geqslant 1 + ax \end{gathered}

\begin{gathered} \sin x \geqslant x - \frac{x^3}{6},\sin x \geqslant x - \frac{1}{2}{x^2}(x > 0) \hfill \\ \tan x > x > \sin x,\left( {0 < x < \frac{\pi }{2}} \right)\\1 - \frac{1}{2}{x^2} \leqslant \cos x \leqslant 1 - \frac{1}{2}{\sin ^2}x \hfill \\ 2\sqrt x > 3 - \frac{1}{x},(x > 1),\tan x < \frac{2}{\pi - 2x},x \in \left( { - \frac{\pi }{2},\frac{\pi }{2}} \right) \end{gathered}

S_{\Delta PBC} \cdot \overrightarrow {AP} + S_{\Delta PCA} \cdot \overrightarrow {BP} + S_{\Delta PAB} \cdot \overrightarrow {CP} = \overrightarrow 0

\text { 重心 } \Leftrightarrow \overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}=\overrightarrow{0}

\overrightarrow{O P}=\overrightarrow{O A}+\lambda(\overrightarrow{A B}+\overrightarrow{A C}), \lambda \in(0,+\infty)

\overrightarrow{O P}=\overrightarrow{O A}+\lambda\left(\frac{\overrightarrow{A B}}{|A B| \sin B}+\frac{\overrightarrow{A C}}{|A C| \sin C}\right)

\text { 垂心 } \Leftrightarrow\overrightarrow{O A} \cdot \overrightarrow{O B}=\overrightarrow{O B} \cdot \overrightarrow{O C}=\overrightarrow{O C} \cdot \overrightarrow{O A}

\overrightarrow{O P}=\overrightarrow{O A}+\lambda\left(\frac{\overrightarrow{A B}}{|A B| \cos B}+\frac{\overrightarrow{A C}}{|A C| \cos C}\right)

\text {内心 } \Leftrightarrow a \cdot \overrightarrow{O A}+b \cdot \overrightarrow{O B}+c \cdot \overrightarrow{O C}=0 \\ \overrightarrow{A O}=\frac{b c}{a+b+c}\left(\frac{\overrightarrow{A B}}{|A B|}+\frac{\overrightarrow{A C}}{|A C|}\right)

\overrightarrow{O P}=\overrightarrow{O A}+\lambda\left(\frac{\overrightarrow{A B}}{|A B|}+\frac{\overrightarrow{A C}}{|A C|}\right)

\overrightarrow{O P}=\overrightarrow{O A}+\lambda\left(\frac{\overrightarrow{A B}}{\sin C}+\frac{\overrightarrow{A C}}{\sin B}\right)

旁心\\\overrightarrow{O P}=\overrightarrow{O A}+\lambda\left(\frac{\overrightarrow{A B}}{|A B|}-\frac{\overrightarrow{A C}}{|A C|}\right)

\cos \left\langle {\overrightarrow {AB} ,\overrightarrow {AC} } \right\rangle = \frac{{{{\overrightarrow {AB} }^2} + {{\overrightarrow {AC} }^2} - {{\overrightarrow {BC} }^2}}}{{2 \cdot |\overrightarrow {AB} | \cdot |\overrightarrow {AC} |}}\\\overrightarrow {AB} \cdot \overrightarrow {AC} = \frac{{{{\overrightarrow {AB} }^2} + {{\overrightarrow {AC} }^2} - {{\overrightarrow {BC} }^2}}}{2}\\ \overrightarrow{A C} \cdot \overrightarrow{B D}=\frac{\left(\overrightarrow{A D}^{2}+\overrightarrow{B C}^{2}\right)-\left(\overrightarrow{A B}^{2}+\overrightarrow{C D^{2}}\right)}{2}

\overrightarrow {AB} \cdot \overrightarrow {AC} = {\overrightarrow {AM} ^2} - {\overrightarrow {CM} ^2} = {\overrightarrow {AM} ^2} - {\left( {\frac{{\overrightarrow {BC} }}{2}} \right)^2}\\ \overrightarrow {AM} \cdot \overrightarrow {CB} = \frac{1}{2}(\overrightarrow {AB} + \overrightarrow {AC} ) \cdot (\overrightarrow {AB} - \overrightarrow {AC} )\\\overrightarrow {BM} = \lambda \overrightarrow {MC} \Rightarrow \overrightarrow {AM} = \frac{1}{{1 + \lambda }}(\overrightarrow {AB} + \lambda \overrightarrow {AC} )

e^x\ge \sum _{i=0}^n \frac{x^i}{i!}(n为偶)\\ e^x\ge \sum _{i=0}^n \frac{x^i}{i!}(n为奇,x\ge0)\\ e^x< \sum _{i=0}^n \frac{x^i}{i!}(n为奇,x<0)\\

\ln x\le \sum _{i=1}^n \frac{(-1)^{i+1} (x-1)^i}{i!}

\left\{ \begin{gathered} \frac{1}{2}\left( {x - \frac{1}{x}} \right) \geqslant \sqrt x - \frac{1}{{\sqrt x }} \geqslant \ln x \geqslant \frac{{2(x - 1)}}{{x + 1}}(x \geqslant 1), \hfill \\ \frac{1}{2}\left( {x - \frac{1}{x}} \right) < \sqrt x - \frac{1}{{\sqrt x }} < \ln x < \frac{{2(x - 1)}}{{x + 1}}(x < 1) \hfill \end{gathered} \right.

{e^\pi } > 23 \Leftrightarrow \pi > \ln 23\\ \because \ln x\le x-1 \\ \ln 2=0.69\\ \begin{array}{l} \therefore\ln 23 &= \ln \frac{{23}}{{25}} \cdot 25\\ &\le (\frac{{23}}{{25}} - 1) + 2\ln 5\\&=2\ln (\frac{{20}}{{19}} \cdot \frac{{19}}{{18}} \cdot \frac{{18}}{{17}} \cdot \frac{{17}}{{16}} \cdot 4)-\frac{2}{{25}}\\&< 2(\frac{1}{{19}} + \frac{1}{{18}}+\frac{1}{{17}} +\frac{1}{{16}} + 2\ln 2)-\frac{2}{{25}}\\ &=2(0.062+0.059+0.056+0.053+2\times0.69)-0.08\\ &=2\times(0.23+2\times0.69)-0.08\\ &=\frac{157}{50}=3.14<\pi \end{array}

$f(x)=e^{x}-\sin x-\cos x, g(x)=e^{x}+\sin x+\cos x$

$x>-\frac{5 \pi}{4} ,$ $f(x) \geq 0$

$g(x) \geq 2+a x$ $a$

$e^x \ge \sin x + \cos x = \sqrt 2 \sin \left( x + \frac{\pi }{4} \right)$

$x \in \left( { - \frac{{5\pi }}{4}, - \frac{\pi }{4}} \right],{e^x} > 0,\sqrt 2 \sin \left( {x + \frac{\pi }{4}} \right) \le 0 \Rightarrow f(x) > 0$

$x \in \left( { - \frac{\pi }{4},\frac{\pi }{4}} \right),f'(x) = {e^x} - \cos x + \sin x,f''(x) = {e^x} + \sqrt 2 \sin \left( {x + \frac{\pi }{4}} \right) > 0$

$\because f'(0) = 0 \Rightarrow f(x) \ge f(0) = 0$

$\left. {x \in \left[ {\frac{\pi }{4}, + \infty } \right.} \right),{e^x} \ge {e^{\frac{\pi }{4}}},\sin x + \cos x \le \sqrt 2 \because {\left( {{e^{\frac{\pi }{2}}}} \right)^{\frac{1}{2}}} > {2^{\frac{1}{2}}} \Rightarrow f(x) > 0$

$e^x \ge \sin x + \cos x \Leftrightarrow F(x) = \frac{{\sin x + \cos x}}{{{e^x}}} \le 1 ,F'(x) = - \frac{{2\sin x}}{{{e^x}}}$

$x \in \left. {\left( { - \frac{{5\pi }}{4},\pi } \right.} \right],F( - \frac{{5\pi }}{4}) = 0,F(0) = 1,F(x) \le 0$

$\left. {x \in \left( \pi , + \infty \right.} \right),f(x) = {e^x} - \sqrt 2 \sin (x + \frac{\pi }{4}) > {e^\pi } - \sqrt 2 > 0$

$e^x$ ${e^x} + \sin x + \cos x \ge ax + 2 \Leftrightarrow h(x) = \frac{{ax + 2 - \sin x - \cos x}}{{{e^x}}} \le 1$

$h\left( { - \frac{\pi }{2}} \right) \le 1 \Rightarrow \frac{\pi }{2}a \ge 3 - {e^{ - \frac{\pi }{2}}} > 2 \Rightarrow a > \frac{4}{\pi } >1$

$h'(x) = \frac{{2\sin x - ax + a - 2}}{{{e^x}}},$ $\varphi (x) = 2\sin x + a(1 - x) - 2,\varphi '(x) = 2\cos x - a$

$a > 2,\varphi '(x) < 0 \Rightarrow \varphi (x) \downarrow$ $\varphi (0) = a - 2 > 0,\varphi (\pi ) = (1 - \pi )a - 2 < 0$

$\exists {x_1} \in (0,\pi )$ $h(x)$ $(0,x_1)\uparrow$ $(x_1,\pi)\downarrow$ $x\in(0,x_1)$ $h(x)>h(0)=1$

$a = 2,\varphi '(x) = 2\cos x - 2 \le 0,\varphi (x) \downarrow ,\varphi (0) = 0$ $h(x)$ $(-\infty,0)\uparrow$ $(0,+\infty)\downarrow$

$h(x)\ge h(0)=1$

$a \in (1,2),\varphi '(x)$ $\left( { - \frac{\pi }{2},0} \right) \uparrow ,\varphi '\left( { - \frac{\pi }{2}} \right) = - a < 0,\varphi '\left( 0 \right) = 2 - a > 0$

$\exists {x_2} \in \left( { - \frac{\pi }{2},0} \right)$ $\varphi(x)$ $\left( { - \frac{\pi }{2},x_2} \right)\downarrow$ $(x_2,0)\uparrow$ $x\in(x_2,0)$ 上

$\varphi(x)<\varphi(0)=a-2<0\Rightarrow h(x)$ $(x_2,0)>h(0)=1$

$x>-\frac{5 \pi}{4}$ $e^x \ge \sin x + \cos x$

${e^x} \ge 2 + ax - {e^x} \Rightarrow {e^x} \ge \frac{a}{2}a + 1 \Rightarrow a = 2$

$a = 2,\varphi '(x) = 2\cos x - 2 \le 0,\varphi (x) \downarrow ,\varphi (0) = 0$ $h(x)$ $(-\infty,0)\uparrow$ $(0,+\infty)\downarrow$

$h(x)\ge h(0)=1$

证法3(必要性探路)

$g(x) \ge 0 \Leftrightarrow F(x) = {e^x} + \sin x + \cos x - ax - 2 \ge 0$

$F(0) = 0 \Rightarrow F'(0) = {e^0} + \cos 0 - \sin 0 - a = 0,a = 2$

$a = 2,\varphi '(x) = 2\cos x - 2 \le 0,\varphi (x) \downarrow ,\varphi (0) = 0$ $h(x)$ $(-\infty,0)\uparrow$ $(0,+\infty)\downarrow$

$h(x)\ge h(0)=1$

导数

分离参数

$f(x)=\frac12ax^2-2ax+\ln x$ $(1,3)$ $a$ 的范围

$<1>$

$f'(x)=ax-2a+\frac1x$ $f'(x)=0$ $x\in(1,3)$ 上有解

$a=\frac{1}{-x^2+2x}$

$-x^2+2x\in(-3,1)$ $a\in(-\infty,-\frac13)\cup (1,+\infty)$

$<2>$

$f'(x)=\frac{ax^2-2ax+1}{x}$ $a=0$ 时，显然不成立

$g(x)=ax^2-2ax+1$ $x=1$

$g(x)$ $(1,3)$ $g(1)\cdot g(3)<0$

$a\in(-\infty,-\frac13)\cup (1,+\infty)$

极值点偏移

$f(x)$ $x=x_0$ $x>x_0$ $x_m=\frac{x_1+x_2}{2}<x_0\rightarrow x_1+x_2<2x_0$ ，这就是极值点偏移，下面我们来看一道例题

$f(x)=(x-2)e^x+a(x-1)^2$ 有两个零点

$f'(x)=(x-1)(e^x+2a)\rightarrow x_1=1,x_2=\ln(-2a)$

$a\in(0,+\infty)$

$x_1<1<x_2$

$x_1+x_2<2\Longleftrightarrow x_1<2-x_2$

$x_2>1\Longleftrightarrow -x_2<-1\Longleftrightarrow 2-x_2<1$

$x_1+x_2<2$ $2-x_2\in(x_1,1)\rightarrow f(2-x_2)<0$

$f(2-x_2)$ $g(x)$ $x>1$

$\begin{cases} g(x)=-xe^{2-x}+a(x-1)^2\\f(x_2)=0\rightarrow(x-2)e^x+a(x-1)^2=0 \end{cases}$

$g(x)=-xe^{2-x}-(x-2)e^x$

$g'(x)=(x-1)e^{2-x}-(x-1)e^x=(x-1)(e^{2-x}-e^x)$

$g(x)$ $(1,+\infty)\downarrow$ $g(1)=0$ $f(2-x_2)<0$ 成立

\text{Proof:}\quad{e^{\sqrt {{x_1} \cdot {x_2}} }} < \frac{{{e^{{x_2}}} - {e^{{x_1}}}}}{{{x_2} - {x_1}}} \quad({x_2} > {x_1})

\Leftrightarrow \frac{{{e^{{x_2} - \sqrt {{x_1} \cdot {x_2}} }} - {e^{{x_1} - \sqrt {{x_1} \cdot {x_2}} }}}}{{{x_2} - {x_1}}} > 1\\ \begin{align} \frac{{{e^{{x_2} - \sqrt {{x_1} \cdot {x_2}} }} - {e^{{x_1} - \sqrt {{x_1} \cdot {x_2}} }}}}{{{x_2} - {x_1}}}&=\frac{{{e^{{x_2} - \sqrt {{x_1} \cdot {x_2}} }} - {e^{{x_1} - \sqrt {{x_1} \cdot {x_2}} }}}}{{({x_2} - \sqrt {{x_1} \cdot {x_2}} ) - ({x_1} - \sqrt {{x_1} \cdot {x_2}} )}}\\ &=\frac{{{e^{{x_2} - \sqrt {{x_1} \cdot {x_2}} }} - {e^{{x_1} - \sqrt {{x_1} \cdot {x_2}} }}}}{{\ln ({e^{({x_2} - \sqrt {{x_1} \cdot {x_2}} )}}) - \ln ({e^{({x_1} - \sqrt {{x_1} \cdot {x_2}} )}})}}\\ &>\sqrt {{e^{{x_2} - \sqrt {{x_1} \cdot {x_2}} + {x_1} - \sqrt {{x_1} \cdot {x_2}} }}}\\ &=\sqrt {{e^{{{(\sqrt {{x_2}} - \sqrt {{x_1}} )}^2}}}}>\sqrt {e^0}=1 \end{align}

\begin{array}{l} 信息熵是信息论中的一个重要概念. \\设 随机变量X所有可能的取值为 1, 2, \\\ldots, n 且P (X=i)=p_{i}>0 (i =\\1, 2, \ldots, n), \sum\limits_{i=1}^{n} p_{i}=1, 定义X的\\信息熵 H(X)=-\sum\limits_{i=1}^{n} p_{i} \log _{2} p_{i} \\ A. 若 n=1 , 则 H(X)=0 \\ B. n=2, 则H (X) 随 p_{1} 的\uparrow而 \uparrow\\ C. 若 p_i=\frac{1}{n}(i=1,2, \ldots, n) \\则H (X) 随着n的增大而增大\\ D. 若n=2m, 随机变量Y所有\\可能的取 值为1 ,2,\ldots, m 且\\P (Y=j)=p_{j}+p_{2m+1-j} \\(j=1,2, \ldots, m) \\则有H (X) \leq H(Y) \end{array}

\begin{array}{l} 0-1 周期序列在通信技术中\\有着重要应用.若序列 a_{1} a_{2} \ldots\\ a_{n} \ldots 满足 a_{i} \in\{0,1\} (i=1\\ ,2 \ldots) 且存在正整数 m,使得 \\ a_{i+m}=a_{i}(i=1,2, \ldots) \\成立,则称其为 0-1 周期序列\\并称满足 a_{i+m}=a_{i}(i=1,2\\ \ldots) 的最小正整数 m 为这个序列\\ 的周期. 对于周期为 m 的 0-1 \\ 序列 a_{1} a_{2} \ldots a_{n}\ldots\\ C(k)=\frac{1}{m} \sum\limits_{i=1}^{m} a_{i} a_{i+k}\\(k=1,2, \ldots, m-1) \\ 是描述其性质的重要指标. 下列\\周期为 5 的 0-1 序列中, 满足\\ C(k) \le \frac{1}{5}(k=1,2,3,4) \\ 的序列是C. 10001 \ldots\\ \end{array}

$f(x) = a\ln x - \sin x + x$ $a$ $f(x)$ $(0, +∞)$ $a$ $\theta \in \left( {\pi ,\frac{{3\pi }}{2}} \right)$ $\cos \theta = 1 + \theta \sin \theta$ ${\theta ^2}\sin \theta < a < 0$ $f(x)$ $(0,2\pi)$ 上恰有两个极值点

$f(x) = (2a{x^2} + bx + 1){e^{ - x}}$ $(1)$ $a = \frac{1}{2}$ $f(x)$ $(2)$ $f(1)=1$ $f(x)=1$ $(0, 1)$ $a$ 的取值范围.

{e^x} > \frac{3}{2}{x^2} - x

f(x) = \frac{{\frac{3}{2}{x^2} - x}}{{{e^x}}} < 1 \Rightarrow f'(x) = \frac{{ - 3{x^2} + 8x - 2}}{{2{e^x}}}

f(\frac{{4 + \sqrt {10} }}{3}) = \frac{{\frac{3}{2}{{\left( {\frac{{4 + \sqrt {10} }}{3}} \right)}^2} - \frac{{4 + \sqrt {10} }}{3}}}{{{e^{\frac{{4 + \sqrt {10} }}{3}}}}}=0.566103

\begin{gathered} g(x) = {e^x} - \frac{3}{2}{x^2} + x \Rightarrow g'(x) = {e^x} - 3x + 1,g''(x) = {e^x} - 3 \hfill \\ g'(\ln 3) = 4 - 3\ln 3 > 0,g'(x) > 0,g(x) > g(0) = 1 \hfill \\ \end{gathered}

f(x) = \frac{{\sin x\cos x}}{{\sin x + \cos x}}\\ f{(x)_{\max }}=\frac{{\sqrt 2 }}{4},x \in \left( { - \frac{\pi }{4},\frac{{3\pi }}{4}} \right)\\ f(\frac{\pi }{4} - x) = f(\frac{\pi }{4} + x)

\begin{gathered} f(x) = {x^2} + x\sin x \hfill \\ g(x) = \left\{ \begin{gathered} \frac{1}{2}x + 1,x \leqslant 0 \hfill \\ \frac{{\ln x + x + 1}}{{x{e^x}}},x > 0 \hfill \\ \end{gathered} \right. \hfill \\ \end{gathered} \\f(g(x)) = m 有四个解 \{ 1 + \sin 1\}

概率

从十个球中取出三个球，放回，求第二次取出球中至少有一个球与第一次取出的三个球中的一个相同的概率

排列组合

P(至少一个一样)=1-P(都不一样)=1-\frac{\mathrm{C}^3_7}{\mathrm{C}_{10}^3}=1-\frac{35}{120}=\frac{17}{24}=70.8\%

反算法

三个全不一样

P(至少一个一样)=1-P(都不一样)=1-\frac{7}{10}\times\frac{6}{9}\times\frac{5}{8}=\frac{17}{24}=70.8\%

直接法

只有一个一样，其余两个不一样

P_1=\frac{3}{10}\times\frac{7}{9}\times\frac{6}{8}\times3=\frac{21}{40}

只有两个一样，其余一个不一样

P_2=\frac{3}{10}\times\frac{2}{9}\times\frac{7}{8}\times3=\frac{7}{40}

三个一样

P_3=\frac{3}{10}\times\frac{2}{9}\times\frac{1}{8}=\frac{1}{120}

P=P_1+P_2+P_3=\frac{17}{24}=70.8\%

三角

\frac{\pi }{2} \le \alpha \le \pi ,\frac{\pi }{2} \le \beta \le \pi\\ \Rightarrow - \pi \le - \beta \le - \frac{\pi }{2}\\ \therefore \frac{\pi }{2} + \frac{\pi }{2} \le \alpha + \beta \le \pi + \pi\\ \frac{\pi }{2} - \pi \le \alpha - \beta \le \pi - \frac{\pi }{2}\\ \left\{ \begin{gathered} \alpha - \beta \in \left[ { - \frac{\pi }{2},\frac{\pi }{2}} \right] \hfill \\ \alpha + \beta \in \left[ {\pi ,2\pi } \right] \hfill \\ \end{gathered} \right.

$\alpha \pm \beta = t$ ${t_{\max }},{t_{\min }}$

\begin{aligned} \sin x+\sin 2 x+\sin 3 x & \leq|\sin x+\sin 2 x+\sin 3 x| \\ &=2|\sin x \cos x+\sin 2 x \cos x| \\ & \leq 2(|\sin x \cos x|+|\sin 2 x \cos x|) \\ & \leq 2 \sqrt{\left(\sin ^{2} x+\cos ^{2} x\right)\left(\cos ^{2} x+\sin ^{2} 2 x\right)} \\ &=2 \sqrt{\frac{25}{16}-\frac{1}{16}(4 \cos 2 x-1)^{2}} \\ & \leq 2 \sqrt{\frac{25}{16}} \\ & \leq \frac{5}{2} \end{aligned}

$a=b=\frac \pi4$

\tan \frac{{\frac{\pi }{4} + \frac{\pi }{4}}}{2} = \tan \frac{\pi }{4} = 1 \ne 0

和差化积

\begin{array}{l} \sin \alpha+\sin \beta=2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}\\ \sin \alpha-\sin \beta=2 \cos \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2}\\ \cos \alpha+\cos \beta=2 \cos \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}\\ \cos \alpha-\cos \beta=-2 \sin \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2} \end{array}

积化和差

\begin{array}{l} \sin \alpha \cos \beta=\frac{1}{2}[\sin (\alpha+\beta)+\sin (\alpha-\beta)] \\ \cos \alpha \sin \beta=\frac{1}{2}[\sin (\alpha+\beta)-\sin (\alpha-\beta)] \\ \cos \alpha \cos \beta=\frac{1}{2}[\cos (\alpha-\beta)+\cos (\alpha+\beta)] \\ \sin \alpha \sin \beta=\frac{1}{2}[\cos (\alpha-\beta)-\cos (\alpha+\beta)] \end{array}

所以

\begin{align} {\text{tan}}\frac{{a + b}}{2} &= \frac{{\sin \frac{{a + b}}{2}}}{{\cos \frac{{a + b}}{2}}}\\ &= \frac{{2 \cdot \sin \frac{{a + b}}{2} \cdot \cos \frac{{a - b}}{2}}}{{2 \cdot \cos \frac{{a + b}}{2}.\cos \frac{{a - b}}{2}}}\\ &=\frac{{\sin \left( {\frac{{a + b}}{2} + \frac{{a + b}}{2}} \right) + \sin \left( {\frac{{a + b}}{2} - \frac{{a + b}}{2}} \right)}}{{\cos \left( {\frac{{a + b}}{2} + \frac{{a + b}}{2}} \right) + \cos \left( {\frac{{a + b}}{2} - \frac{{a + b}}{2}} \right)}} \\ &=\frac{{\sin a + \sin b}}{{\cos a + \cos b}} \end{align}

均值

\begin{aligned} f(a, b, c): &=a^{2}+b^{2}+c^{2}+2 a b c \\ &=(a+b+c)^{2}-2(a b+b c+c a)+2 a b c \\ &=1-2 a b(1-c)-2(a+b) c \\ &=1-2 a b(1-c)-2(1-c) c \\ &=1-2(1-c)(a b+1-a-b) \\ &=1-2(1-a)(1-b)(1-c) . \end{aligned}

1>f(a, b, c) \geq 1-2\left[\frac{(1-a)+(1-b)+(1-c)}{3}\right]^{3}=\frac{11}{27}

解几

$\text{Rt}\triangle ABC$ $\angle A=\frac\pi2$ $AM$ $AD$ 为角平分线

$\vec{AB}$ $x$ $\vec{AC}$ $y$ $AB=a$ $AC=b$

由中点公式

\left\{ \begin{gathered} {x_M} = \frac{{{x_B} + {x_C}}}{2} \hfill \\ {y_M} = \frac{{{y_B} + {y_C}}}{2} \hfill \\\end{gathered} \right.\Rightarrow M\left( {\frac{a}{2},\frac{b}{2}} \right)

$BC$ $\frac{x}{a} + \frac{y}{b} = 1$ $y=x$ 联立得

D\left( {\frac{{ab}}{{a + b}},\frac{{ab}}{{a + b}}} \right)

\begin{align} \therefore A{M^2} &= \frac{{{a^2}}}{4} + \frac{{{b^2}}}{4}\\ &={\sqrt {\frac{{{{\left( {\frac{a}{{\sqrt 2 }}} \right)}^2} + {{\left( {\frac{b}{{\sqrt 2 }}} \right)}^2}}}{2}} ^2} \\AD^2&=2{\left( {\frac{{ab}}{{a + b}}} \right)^2}\\ &= {\left( {\frac{2}{{\frac{{\sqrt 2 }}{a} + \frac{{\sqrt 2 }}{b}}}} \right)^2} \end{align}

已知

\begin{array}{c} H_{n}=\frac{n}{\sum \limits_{i=1}^{n}\frac{1}{x_{i}}}= \frac{n}{\frac{1}{x_{1}}+ \frac{1}{x_{2}}+ \cdots + \frac{1}{x_{n}}} \\ Q_{n}=\sqrt{\sum \limits_{i=1}^{n}x_{i}^{2}}= \sqrt{\frac{x_{1}^{2}+ x_{2}^{2}+ \cdots + x_{n}^{2}}{n}} \\ H_{n}\leq Q_{n} \end{array}

$H_2\le Q_2$

得

AM\ge AD

$AB=AC$ 时成立

另解，由中线定理

{\left( {2AM} \right)^2} + B{C^2} = 2(A{B^2} + A{C^2}\\AM = \frac{{2({a^2} + {b^2}) - ({a^2} + {b^2})}}{4} = \frac{{{a^2}}}{4} + \frac{{{b^2}}}{4}\\

$S_{\triangle ABC}=S_{\triangle ABD}+S_{\triangle ACD}$

\frac{1}{2}AB \cdot AC = \frac{1}{2}AB \cdot AD \cdot \sin \frac{\pi }{4} + \frac{1}{2}AC \cdot AD \cdot \sin \frac{\pi }{4}\\ AD = \frac{{\sqrt 2 AB \cdot AC}}{{AB + AC}} = \frac{{\sqrt 2 ab}}{{a + b}}

$F_1$ 为极点，垂直于左准线方向为极轴建系

$\rho = \frac{{ep}}{{1 - e\cos \theta }}$

L = \rho _\theta + \rho _{\pi - \theta} = \left| \frac{2ep}{ {1 - {e^2}\cos ^2}\theta } \right|

\begin{aligned} \therefore {S_{PMQN}}& = \frac{1}{2}|PQ| \cdot |MN|\\ & = \frac{1}{2} \cdot \frac{{2ep}}{{1 - {e^2}{{\cos }^2}\theta }} \cdot \frac{{2ep}}{{1 - {e^2}{{\cos }^2}(\pi-\theta)}}\\ &=\frac{{8{e^2}{p^2}}}{{4 - 4{e^2} + {e^4}{{\sin }^2}2\theta }} \end{aligned}\\ \because e=\frac{c}{a},p=\frac{{{a^2}}}{c} - c = \frac{{{b^2}}}{c} \\\therefore {S_{{\text{min}}}}=\frac{{8{e^2}{p^2}}}{{4 - 4{e^2} + {e^4}}}=\frac{{8{a^2}{b^4}}}{{{{({a^2} + {b^2})}^2}}} \\{S_{{\text{max}}}} = \frac{{2{e^2}{p^2}}}{{1 - {e^2}}} = 2{b^2}

另解

$PQ$ $x=my-c$ ，联立

\begin{cases} \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\\ x=my-c \end{cases} \Rightarrow ({a^2} + {b^2}{m^2}){y^2} - 2mc{b^2}y - {b^2}{c^2} = 0

|PQ| = \sqrt {1 + {m^2}} \cdot |{y_1} - {y_2}| = \frac{{2a{b^2}({m^2} + 1)}}{{{a^2} + {b^2}{m^2}}}

$- \frac{1}{m}$ $m$

|MN| = \frac{{2a{b^2}\left[{{\left( - \frac{1}{m}\right)}^2} + 1\right]}}{{{a^2} + {b^2}{{( - \frac{1}{m})}^2}}} = \frac{{2a{b^2}\left({m^2} + 1\right)}}{{{a^2}{m^2} + {b^2}}}\\

$t=m^2$

\begin{aligned} \therefore {S_{PMQN}}& = \frac{1}{2}|PQ| \cdot |MN|\\ & = \frac{{2{a^2}{b^4}{{({t} + 1)}^2}}}{{({a^2} + {b^2}{t})({a^2}{t} + {b^2})}}\\ &=\frac{{2{a^2}{b^4}({t^2} + 2t + 1)}}{{{a^2}{b^2}{t^2} + ({a^4} + {b^4})t + {a^2}{b^2}}} \end{aligned}\\

$t=0\Rightarrow S= 2{b^2}$

$t \ne 0$ ，分离常数

\begin{aligned} {S_{PMQN}}& = 2{a^2}{b^4} \cdot \left[ {\frac{{\left( {2 - \frac{{{a^4} + {b^4}}}{{{a^2}{b^2}}}} \right)t}}{{{a^2}{b^2}{t^2} + ({a^4} + {b^4})t + {a^2}{b^2}}} + \frac{1}{{{a^2}{b^2}}}} \right] \\&=2{b^2} - \frac{{2({a^2} - {b^2}){b^2}}}{{{a^2}{b^2}\left( {t + \frac{1}{t}} \right) + {a^4} + {b^4}}} \\&\ge 2{b^2} - \frac{{2({a^2} - {b^2}){b^2}}}{{{{({a^2} + {b^2})}^2}}} \\&=\frac{{8{a^2}{b^4}}}{{{{({a^2} + {b^2})}^2}}} \end{aligned}

$t=m^2=1,k=\pm 1$ 时取等

\because\lim_{t \to 0/+\infty} \frac{{2({a^2} - {b^2}){b^2}}}{{{a^2}{b^2}\left( {t + \frac{1}{t}} \right) + {a^4} + {b^4}}}=0

\\ \therefore {S_{{\text{min}}}}=\frac{{8{a^2}{b^4}}}{{{{({a^2} + {b^2})}^2}}} \\{S_{{\text{max}}}} = 2{b^2}

注

\frac{{2{a^2}{b^4}{{({t} + 1)}^2}}}{{({a^2} + {b^2}{t})({a^2}{t} + {b^2})}}\ge\frac{{2{a^2}{b^4}{{(t + 1)}^2}}}{{\frac{1}{4} \cdot {{[({a^2} + {b^2})t + ({a^2} + {b^2})]}^2}}}=\frac{{8{a^2}{b^4}}}{{{{({a^2} + {b^2})}^2}}}

代数

pq = \frac{{{{(p + q)}^2} - {{(p - q)}^2}}}{4}

\frac{1}{{p + 1}} - \frac{1}{{p - 1}} = \frac{2}{{{p^2} - 1}} \Rightarrow {p^2} = 1 + \frac{2}{{\frac{1}{{p + 1}} - \frac{1}{{p - 1}}}}

f(x)=\frac{2.02 \times 10^{6}+x+\sin x+2020^{0}-\lg 2-\lg 5}{2} \\x\in[-520,520] ,f(x)_{\text{max}}=L,f(x)_{\text{min}}=O \\g(x)=\frac{2020^{0}+e^{\ln 520}+(-1)^{3}+x^{3}+\lg \frac{2021-x}{2021+x}}{2} \\x\in[2020,2020] ,g(x)_{\text{max}}=V,g(x)_{\text{min}}=E

1988 \quad\text{IMO#6} \\a 和 b 是是正整数, 且 a b+1 整除 a^{2}+b^{2} . \\求证 \frac{a^{2}+b^{2}}{a b+1} 为完全平方数. \\韦达跳越 (\text{Vieta jumping})

\tan x + \frac{2}{{2x - \pi }} < 0(x \in \left[ {0,\frac{\pi }{2}} \right))

F = k\frac{{{q_1}{q_2}}}{{{r^2}}}

GM = g{R^2}\\M = \frac{{4{\pi ^2}{r^3}}}{{G{T^2}}}

{P_t} = F{v_t}\cos \theta \\ \bar P = \frac{W}{{\Delta t}} = F\bar v\cos \theta

\left\{ \begin{gathered} r = \frac{{mv}}{{Bq}} = \frac{1}{B}\sqrt {\frac{{2mU}}{q}} \hfill \\ T = \frac{{2\pi m}}{{Bq}} = \frac{{2\pi r}}{v} \hfill \\ t = \frac{{\theta m}}{{Bq}} = \frac{{\theta r}}{v} \hfill \\ \end{gathered} \right.

电场

\left\{ \begin{gathered} \varphi = \frac{{{E_p}}}{q} (能) \\ E = \frac{F}{q} (力) \\ \end{gathered} \right.

{U_{AB}} = {\varphi _A} - {\varphi _B} = \frac{{{W_{AB}}}}{q} = \frac{{{E_{pA}} - {E_{pB}}}}{q} = Ed\cos \theta

C = \frac{Q}{U} = \frac{{{\varepsilon _r}S}}{{4\pi kd}}

恒定电流

I = \frac{{\Delta q}}{{\Delta t}} = nesv = \frac{U}{R} = \frac{E}{{R + r}}

E = \frac{W}{q} = I(R + r) = U_内(Ir) + U_外

\left\{ \begin{gathered} {R_I} = \frac{{{R_g}}}{{n - 1}} \hfill \\ {R_V} = (n - 1){R_g} \hfill \\ \end{gathered} \right.

\begin{gathered} Q = {I^2}Rt \hfill \\ P = IU=\frac{{{U^2}}}{R} = {I^2}R \hfill \\ \end{gathered}

R = \rho \frac{l}{S}

W(正功负功)\\{U_{AB}} = {\varphi _A} - {\varphi _B}\\ W_{AB}=-\Delta E_电\\ \Delta E_p\quad\Delta E_k

$A$ $t_1,t_2$

\left\{ \begin{gathered} {v_{A1}} = g\sin \theta {t_1} \hfill \\ {v_{B1}} = a{t_1} \hfill \\ \end{gathered} \right.

最终速度

\left\{ \begin{gathered} {v_{A2}} = g\sin \theta {t_1} \hfill \\ {v_{B2}} = a({t_1} + {t_2}) \hfill \\ v_{A2}=v_{B2} \end{gathered} \right.

能追上

{x_B} = \frac{{{v_{B2}}}}{2}({t_1} + {t_2}) = {v_{A2}}{t_2} \Rightarrow {t_1} = {t_2}

a > \frac{{g\sin \theta {t_1}}}{{{t_1} + {t_2}}} = \frac{{g\sin \theta }}{2}

另

\begin{gathered} \frac{1}{2}a{({t_1} + {t_2})^2} = g\sin \theta {t_1}{t_2} \hfill \\ a{t_1}^2 + 2{t_2}(a - g\sin \theta ){t_1} + a{t_2}^2 = 0 \hfill \\ \Delta = 4{t_2}^2({a^2} - 2ag\sin \theta + {g^2}{\sin ^2}\theta ) - 4{a^2}{t_2}^2 > 0 \hfill \\ \end{gathered}

\begin{cases}\rm _{92}^{238}U+ _{0}^{1}n\rightarrow_{56}^{144}Ba+_{36}^{89}Kr+3_{0}^{1}n\\ \rm_{1}^{2}H+ _{1}^{3}H\rightarrow_{2}^{4}He+_{0}^{1}n+17.6MeV \end{cases}\\ \Delta E = \Delta m{c^2},m \to \frac{\rm{MeV}}{{{c^2}}}(质量亏损)\\核力\to强相互作用,短程(1.5\times10^{-15})\\弱力(\beta 衰变,质子-中子转变)10^{-18}\rm m\\ 万有引力<电磁力(10^{35})<弱力<强力

可逆反应 \\ \ce{ N2 +3H2 <=>[高温][催化剂] 2 NH3\\ Fe^{3+}<=>Fe(SCN)3\\ NO2(红棕)<=>N2O4\\ Cr2O7^2-<=>CrO4^-\\ HA<=>H+ +A- \\ BOH<=>B+ +OH- \\ AB(s)<=>A+ +B- \\ 2SO2 +O2<=>[ 催化剂]2SO3\\ Cl2<=>HClO + Cl- \\ C +CO2<=>2CO \\ 2HI<=>[\triangle]H2 +I2 }\\酯化,酸性水解

还原性\\ \ce{ 金属单质( Na,K,Zn,Al) \\ Cl- (MnO4^-) ->Cl2 \\ S^2-,H2S,HS- -> S \\ NO2 ,NO2- ->NO3- \\ SO3^2- ,HSO3- ->SO4^2-\\ NH3->NO->NO2\\ C2O4^2- (有机)->CO2 \\- CHO->-COOH \\ (能得氧去氢)H2,CO,C \\ NaBH4,LiAlH4, SO2\\ Fe^2+,Cu+,Cr^3+ }

氧化剂\\ \ce{ Cr2O7^2-,MnO4- \\ O2(空气),O3 \\ NO3-,NO2- \\ SO2->S,MnO2\\ ClO3-,ClO- \\ Br2,Cl2,I2,F2\\ Na2O2,H2O2\\ Fe^3+,Cu^2+ \\ H+ ,浓 H2SO4 }

漂白剂\\ \ce{ 氧化性Ca(ClO)2, O3\\ HClO,H2O2,NaClO\\ HNO3,NO2\\ 化合型SO2,Na2S2O6\\ 吸附型C木炭或活性碳 }

N_\text{A}\\ \ce{ 石墨->\frac32 C-C\\ ->\frac12 六元环\\ 金刚石->2 C-C\\ P4->6P-P\\ S8->8S-S \\ SiO2->4Si-O\\ Si->2Si -Si\\ 氢键H2O->2N_A\\ NH3,HF->N_A }

\ce{ AgCl白S黄Cu(OH)2蓝\\ AgBr淡黄C黑AgI黄色\\ CuO,MnO2,Ag2O黑\\ Fe3[Fe(CN)6]蓝\\ Fe(OH)2白,灰绿,红褐\\ Fe(OH)3红褐FeS2黄色\\ BaSO4,CaCO3白\\ Cu2O砖红Fe2O3红色\\ Fe3O4黑有磁FeO\\ Mg(OH)2,Al(OH)3白\\ ZnS,FeS可溶于酸\\ CuS,PbS氧化性酸\\ CaSO4,Ca(OH)2微溶\\ Cu2(OH)2CO3暗绿色\\ }

\ce{ Cu^2+ 少黄中绿多蓝I2紫黑固体\\ Fe^3+ 黄 Fe^2+ 浅绿Br2红棕色\\ \\ NO2红棕,N2O4无O2复燃\\ Cl2黄绿F2浅黄绿HCl白雾\\ \\ NO无色(O2管口)变为红棕色\\ SO2品红,SO3(BaCl2)\\ H2S(湿Pb(Ac)2)黑(还原性)\\ CrO4- 黄 Cr2O7^2- 橙黄 \\ MnO4- 紫红Cr^3+ 灰绿\\ \rho (CCl4)>\rho (H2O)无色液体\\ Br2橙红,I2紫红(淀粉溶液)\\ H+/OH- 酸碱指示剂(石蕊)\\ CO3^2- (CO2)Ca(OH)2\\ SO3^2-(SO2)品红褪色\triangle恢复 \\ HCO3-/HSO3- (CaCl2)无 \\ SO4^2- 先HCl再BaCl2白沉\\ Cl- AgNO3(HNO3)白沉\\ NH4+ (浓NaOH,\triangle)红石蕊 }

化学反应相关量的计算

首先应明确问的是什么

在草稿纸上写出要求的量(目标)

写出化学方程式

根据三段式(关系式)列式(方程)计算

\Delta {\text{H}}

明确要求的反应是什么

写出其热化学方程式(标明状态)

看已知哪些量如反应热

燃烧热活化能能量

看特殊产物注意系数和左右

写出表达式再代入数据计算

方程式的书写

明确方程式类型离子化学有机可逆

看所处环境酸性碱性

依据所给条件(过量)写出反应物及产物

并标明沉淀气体反应条件

看得失电子电荷守恒原子守恒配平

最后检验是否科学

电化学

明确是电解池还是原电池(自发性)

看反应环境氧化物碳酸盐酸性碱性

具题意及得失电子找到反应物产物阴阳极

明确电池工作过程即离子移动

具体分析溶液相关状态量的变化

计算具体细节(方程式)

写出方程式(标明变价元素化合价和转移电子)

\mathop {\mathop {{\text{C}}{{\text{H}}_4}}\limits_{ \uparrow 1 \times 8} }\limits^{ - 4\;\;\;\;} + 2\mathop {\mathop {{\text{NO}}}\limits_{ \downarrow 2 \times 2} }\limits^{ + 2\;\; - 2} + \mathop {\mathop {{{\text{O}}_2}}\limits_{ \downarrow 2 \times 2} }\limits^0 \rightleftharpoons \mathop {{\text{C}}{{\text{O}}_2}}\limits^{ + 4\;\; - 2} + \mathop {{{\text{N}}_2}}\limits^0 + 2\mathop {{{\text{H}}_2}{\text{O}}}\limits^{\;\;\;\; - 2} {\text{(l)}}

$\times$ 化合价变化量)

通过电荷守恒质量守恒得失电子守恒得到关系

写出方程式,左侧写上物理量的符号,依照起始和变化

\begin{gathered} \quad \quad {\text{CO}} + 2{{\text{H}}_2} \rightleftharpoons {\text{C}}{{\text{H}}_3}{\text{OH}} \hfill \\ n:3 - \underline t \;2 - \underline {2t} \quad 0 + \underline t ({\text{He}}:1) \hfill \\ V:3 \to 2.4(6 \to 6 - 2t) \Rightarrow t = 0.6 \hfill \\ \alpha ({\text{CO}}) = \frac{t}{3} = 0.2,p({{\text{H}}_2}) = \frac{{2 - 2t}}{{6 - 2t}}{p_0} \hfill \\ \end{gathered}

利用相似性成比例,根据题给条件,得到相关数据

最后列出表达式进行计算(三段式,起转终)

${\text{C}}{{\text{l}}_2} \sim {\text{FeC}}{{\text{l}}_{2/3}} \sim {e^ - } \sim 2{\text{C}}{{\text{l}}^ - }$

明确所要研究的体系中的物质

找出他们之间的比例关系

利用电子转移和原子守恒列式计算

尤其注意始态终态(终态法，忽略中间)

离子平衡/沉淀溶解

${\text{HX}}$ 为何)

相互作用的过程是什么(定性分析)

{\text{(1mol}} \cdot {{\text{L}}^{ - 1}}{\text{)NaOH}} \to {{\text{H}}_2}{\text{A}}(c?V=20{\text{ml}})

通过特殊点特殊数据找关系

${{\text{K}}_w},{{\text{K}}_a},{{\text{K}}_b},{{\text{K}}_h},{Q_c}{\text{,}}{{\text{K}}_{{\text{sp}}}}$ 进行定量分析

\ce{Ag+＞Hg^2+＞Fe^3+＞Cu^2+ ＞H+＞Pb^2+＞Sn^2+＞Fe^2+＞Zn^2+\\ > H+(H2O)>\underline{\ce{Al^3+ >Mg^2+>Na+>Ca^2+>K+}}(只在熔融时)}\\ 失电子(阳极):\ce{活性电极>S^2-＞I^-＞Br^-＞Cl- ＞OH^-＞}含氧酸根离子\\ \ce{ H He Li Be B C N O F Ne Na Mg Al Si P S Cl Ar K Ca}\\\ce{ Sc Ti V Cr Mn Fe Co Ni Cu Zn}

有机反应及官能团总结

$\ce{O2}$ 反应)

$=,\equiv,\ce{-CHO}$ ，酮,苯

$\ce{HX,H2O,X(H)2}$ $1,2/1,4-$ 加成(竞争反应，取决条件)

$\ce{R－L +A-B→R－A +L-B}$ (硝化，酯化)

$\ce{-X +NaOH->[H2O][\triangle]-OH +NaX }$ $\ce{HO-NO2->[浓硫酸][\triangle]}$

$\ce{NaOH->[醇][\triangle]不饱和物+NaX }$ $\ce{->[浓硫酸][170^\circ C]}$ 脱去小分子

$\ce{->[浓硫酸][140^\circ C]}$ $\ce{-CH2OH->-CHO->-COOH}$

$\ce{Br2}+$ $\ce{FeCl3}$ 紫色反应

$\ce{Ag(NH3)2OH/Cu(OH)2}$ $\ce{-CHO}$ )

$\ce{R-CO-\boxed{\ce{OH +H}}-O-R'<=>[浓硫酸][\triangle]R-COO-R' +\underline{\ce{H2O}}}$

$\ce{<=>[稀硫酸][\triangle]}$ $\ce{->[\triangle]}$ 不可逆(油脂：皂化反应)

$\underset{淀粉}{\underline{\ce{(C6H10O5)}_n}}+n\ce{H2O->[酸/酶]}n\underset{葡萄糖}{\underline{\ce{C6H12O6}}}$

$\ce{-COOH +H2N -->-CONH - +H2O }$

$\ce{->}$ $\ce{->}$ 缩聚(缩聚物)

$\ce{->[H+]}$ $\ce{->[OH-]}$ 网状酚醛树脂

$\ce{a A +b B<=> pC +qD}$

速率常数

$v_正 = k_正 \cdot {c^a}({\text{A}}) \cdot {c^b}({\text{B}})\\ v_逆 = k_逆 \cdot {c^p}({\text{C}}) \cdot {c^q}({\text{D}})\\$

$v_正=v_逆$ ${\text{K}} = \frac{{{c^p}({\text{C}}) \cdot {c^q}({\text{D}})}}{{{c^a}({\text{A}}) \cdot {c^b}({\text{B}})}} = \frac{k_正}{k_逆}$

分压平衡常数

${{\text{K}}_p} = \frac{{{p^p}({\text{C}}) \cdot {p^q}({\text{D}})}}{{{p^a}({\text{A}}) \cdot {p^b}({\text{B}})}}$

${{\text{K}}_p} = \frac{{{\varphi ^p}({\text{C}}) \cdot {\varphi ^q}({\text{D}})}}{{{\varphi ^a}({\text{A}}) \cdot {\varphi ^b}({\text{B}})}} \cdot {p_0}^{\frac{{c \cdot d}}{{a \cdot b}}}$

${{\text{K}}_n} = \frac{{{\varphi ^p}({\text{C}}) \cdot {\varphi ^q}({\text{D}})}}{{{\varphi ^a}({\text{A}}) \cdot {\varphi ^b}({\text{B}})}}$

转化率

$\alpha = \frac{n_{参加反应的物质}}{n_{投入的物质}}\times100\%$

原子利用率

$\frac {m_{期望产物}}{m_{生成物}}\times100\%$

纯度

$\frac {m_{某种物质}}{m_{混合物}}\times100\%$

产率

$\frac {m_{实际产量}}{m_{理论产量}}\times100\%$

\ce{CO(g) +H2O(g) <=>CO2(g) +H2(g)\quad\Delta H=- 41 kJ/mol\quad}

\ce{H+ +OH- = H2O \quad\Delta H=- 57.3 kJ/mol}

\ce{N2(g) +3H2(g) <=>[高温][催化剂] 2 NH3(g) \quad\Delta H=- 92.4 kJ/mol}

\ce{2 NO +2CO <=>2CO2 +N2}

\ce{CO2(g) +3H2(g)<=>CH3OH(l) +H2O(l)\quad\Delta H=- 93.8 kJ/mol}

\ce{CO +2H2<=>CH3OH }

{k_1} < - {k_2} \Rightarrow {k_1} + {k_2} < 0\\设T' > T \Rightarrow{\frac{1}{{T'}}< \frac{1}{T}},\frac{1}{{T'}} - \frac{1}{T} < 0\\\frac{{\lg {{\text{K}}_{{\text{p}}2}}' - \lg {{\text{K}}_{{\text{p}}2}}}}{{\frac{1}{{T'}} - \frac{1}{T}}} + \frac{{\lg {{\text{K}}_{{\text{p}}1}}' - \lg {{\text{K}}_{{\text{p}}1}}}}{{\frac{1}{{T'}} - \frac{1}{T}}} < 0\\\lg \frac{{{{\text{K}}_{{\text{p}}2}}^\prime }}{{{{\text{K}}_{{\text{p}}2}}}} \cdot \frac{{{{\text{K}}_{{\text{p}}1}}^\prime }}{{{{\text{K}}_{{\text{p}}1}}}} > 0 \Rightarrow {{\text{K}}_{{\text{p}}1}}^\prime {{\text{K}}_{{\text{p2}}}}^\prime > {{\text{K}}_{{\text{p}}1}}{{\text{K}}_{{\text{p}}2}}

\text{Van't Hoff equation}\\ \Delta G^{\ominus}=\Delta H^{\ominus}-T \Delta S^{\ominus} \\ \Delta G^{\ominus}=-R T \ln K \\ \Rightarrow\ln K=-\frac{\Delta H^{\ominus}}{R T}+\frac{\Delta S^{\ominus}}{R} \\ \frac{\text{d} \ln K}{\text{d} \frac{1}{T}}=-\frac{\Delta H^{\ominus}}{R}

\text{Arrhenius equation}\\ k = A{e^{ - {E_a}/RT}}\\R\ln k = - {E_a} \cdot \frac{1}{T} + C\\ a{\text{A}} + b{\text{B}} \to c{\text{C}}\\ v = \frac{1}{a}\frac{{{\text{d}}({\text{A}})}}{{{\text{d}}t}} = k{({\text{A}})^a}{({\text{B}})^b}\\ \int\limits_{{{({\text{A}})}_0}}^{({\text{A}})} {\frac{{{\text{d}}({\text{A}})}}{{({\text{A}})}} = - \int\limits_0^t {k{\text{d}}t} } \\ \Rightarrow \lg \frac{{({\text{A}})}}{{{{({\text{A}})}_0}}} = - \frac{{kt}}{{2.30}}

以句子为基本单位来审题

每读完一个句子就进行思考

圈出题眼和关键条件，数据

对图像表格注意一看轴二看线

$x$ $y$ (因变量)

要有序,先找到切入点再深入

对一道题要有关联性,全局意识

不要忘记要求(写结果,有效数字)

试卷的使用:圈点勾画关键条件

$F=m_1g$

$ma=kx-mg\sin\theta$

也可写出难题的大致框架思路

不要再试卷上计算化简，画图用铅笔

进行圈出不确定的以便后续检查和验算

草稿纸使用:所有计算和化简

没有图的题用铅笔画图(复杂的画大点)

$\boxed{必须}$ 在草稿纸上写

大题在上写出已知条件找多种方向突破

$v = \frac{{\bcancel{m}{v_0}}}{{\sqrt 2 \bcancel{m}}}=v_0$

${({x^2} + {y^2})_{{\text{min}}}}$

计算化简等式

左右变化移动

\begin{gathered} a = c \Rightarrow a \pm b = c \pm b \hfill \\ a + b = c \Rightarrow a = c - b \hfill \\ \end{gathered}

交叉相乘

\frac{a}{c} = \frac{b}{d} \Rightarrow ad = bc

左右同乘除(同上同下)

\frac{a}{c} = \frac{b}{d}\Rightarrow\frac{am}{cn} = \frac{bm}{dn}

分数

\frac{a}{c} = \frac{{a \cdot t}}{{c \cdot t}} = \frac{{b /t}}{{d /t}}\

先打大括号再乘开/计算

- {(2x - 1)^2} = - (4{x^2} - 4x + 1) = - 4{x^2} + 4x - 1

分别算前算后再整合

\begin{gathered} 30 \times 800 = 3 \times 8 \times {10^3} = 24 \times {10^3} = 2.4 \times {10^4} \hfill \\ 5 \times {10^4} + 3 \times {10^5} = 0.5 \times {10^5} + 3 \times {10^5} = 3.5 \times {10^5} \hfill \\ \end{gathered}

计算(由浅入深)目标，单位

\begin{gathered} c({{\text{I}}^ - }) = \frac{{m({\text{mg}})}}{{V({\text{L}})}} = \frac{{n \cdot {\text{Mr}}}}{V} = \frac{{2 \cdot n({\text{C}}{{\text{u}}^{2 + }}) \cdot {\text{Mr}}}}{V} \hfill \\ = \frac{{2 \cdot 2.5 \times {{10}^{ - 3}} \cdot {{10}^3}({\text{g}} \to {\text{mg)}} \cdot 127}}{{2.5}} = 254\boxed{{\text{mg}} \cdot {{\text{L}}^{ - 1}}} \hfill \\ \end{gathered}

有效数字

r(.3 小数\to /3 \to有效数字 ) = {k_{U - I}} = \frac{{1.4 - 1.2}}{{0.2 - 0.1}} = 2.00\Omega

边缘计算不忽略

S = \boxed{6\sqrt 3} \frac{k}{{1 + {k^2}}} = \frac{\boxed{6\sqrt 3}}{{1/k + k}} \leqslant \frac{\boxed{6\sqrt 3}}{2} = 3\sqrt 3

\frac{{{k_1} - {k_2}}}{{{k_2} - {k_3}}} = \frac{{{k_1} - {k_3} + {k_3} - {k_2}}}{{{k_2} - {k_3}}} = \frac{{{k_1} - {k_3}}}{{{k_2} - {k_3}}} - 1 = - 1 \underline{- 1} = - 2

最后带入数据(对照)

\begin{gathered} t = \frac{{3\lg 2 - \lg 15}}{{2\lg 2 - \lg 5}} = \frac{{4\lg 2 - \lg 3 - 1}}{{3\lg 2 - 1}} = \frac{{4 \cdot 0.301 - 0.477 - 1}}{{0.903 - 1}} \approx 2.8\hfill \\ \hfill \\ \end{gathered}

x=\underset{振幅}{\underline{A }}\sin\left( \underset{角速度}{\underline{\omega }}t+\underset{初相}{\underline{\varphi_{0}}}\right)=A \sin \left(\underset{相位}{\underline{\frac{2 \pi}{T} t+\varphi_{0}}}\right)

T=\sqrt {\frac{l}{g}} ，v = \frac{\lambda }{T} = \lambda f,F = - \frac{{mg}}{l}x,f=\frac{1}{2 \pi \sqrt{L C}}

n = \frac{{\sin i}}{{\sin r}},\sin C = \frac{1}{n},v = \frac{c}{n},T=2 \pi \sqrt{L C} \quad

{r_2} - {r_1} = n\lambda ,x=n\frac{l}{d}\lambda,\Delta x = \frac{l}{d}\lambda

暗纹：\;{r_2} - {r_1} = (2n - 1)\frac{\lambda }{2}

L = {L_0}\sqrt {1 - \frac{{{v^2}}}{{{c^2}}}} ,m = \frac{{{m_0}}}{{\sqrt {1 - \frac{{{v^2}}}{{{c^2}}}} }}

物理做题流程

明确考察内容

$F=BIL=\frac{B^2L^2v}{R+r}=Bqv$

找出研究对象

$M$ $q$ $v_0$

看外界条件

$\mu mg\cos\theta$ $mg$ ，斜面是否固定，电场和磁场有无边界，是轻绳还是杆，注意划出关键点

借助图像进行受力分析,过程分析,电路分析，轨迹分析

对研究对象分析受力，选择合成分解方向，注意不多不少，平衡力不要省略

重力，弹力，摩擦力，拉力，电场力，洛伦兹力，静电力，万有引力，外力等

将等效电路绘制出，各个过程的对象及运动过程画出，并标上数据

按题目所问列方程

最好由浅入深，一步一步列方程，不要进行计算，方程列对了也可以得分

方程列完后进行计算并评估所得结果

看所得结果是否符合常识

易算不贪快

难算不心慌

小算不图便

大算不着急

繁算不厌烦

算一步对一步

步步对一遍过

受力分析细节(场力及接触力)

$\vec F = \left\{ \begin{gathered} \vec Eq \hfill \\ Uq/d \hfill \\ \end{gathered} \right.$ $+ q/ - q$ 场强方向

$\overrightarrow F = \overrightarrow B \times \overrightarrow v q,\left\{ \begin{gathered}电性+q/ - q \hfill \\粒子的速度 \overrightarrow v \hfill \\ 磁场方向\overrightarrow B (\sin\theta) \hfill \\ \end{gathered} \right.$

$\overrightarrow F = \overrightarrow B \times \overrightarrow I L$ 结合电路分析电流

$F = G\frac{{Mm}}{{{r^2}}}$ 方向指向另一个天体

$mg$ 在斜面上经常分解成2个力

$F = k\frac{{Qq}}{{{r^2}}}$

$f\left\{ \begin{gathered} \hfill静：方向大小由供需决定 \\ 动：和相对运动方向反\mu {F_N} \hfill \\ \end{gathered} \right.$

压力：垂直接触面(若固定约束力)

$\left\{ \begin{gathered} 弹簧弹力 F = k\Delta x(形变量l-l_0) \hfill \\ 杆的弹力，方向任意，由供需决定 \end{gathered} \right.$

$\boxed{收缩}$ 方向，区分死结活结

$\boxed{\sum F=0,a=0}$

复合场

$+ q/ - q$ ，计不计重力

看初速度，分析受力，注意方向

一找圆心二找半径，电场找抛体轨迹

带入数据和题中条件标上数据

$T$ $t$ $v_t$

利用圆周运动和运动学能量多角度求解

注意多解性和检验是否科学合理

\begin{gathered} f_动 = \mu {F_N} \hfill \\ F_{库仑} = k\frac{{{q_1}{q_2}}}{{{r^2}}} \hfill \\ F_安 = BIL \hfill \\ F_电 = Eq = \frac{{Uq}}{d} \hfill \\ F_洛 = Bqv \hfill \\ F_n = m\omega v \hfill \\ G = mg \hfill \\ F_弹 = k|l - {l_0}| \hfill \\ F_引 = G\frac{{mM}}{{{r^2}}} \end{gathered}

E_{pn}=-k\frac{e^2}{r_n} \\E_{kn}=k\frac{e^2}{2r_n} \\ E_n=-k\frac{e^2}{2r_n}

位移\vec{x}\\ 速度\vec{v}\\ 加速度\vec{a}\\ 电场强度\vec{E}\\ 磁感应强度\vec{B}\\ 冲量\vec{I}\\ 动量\vec{p}\\ 变化量\vec{\Delta v}\\