用户:施馨柠

分数

$\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{4}{5}, \frac{5}{6}$

根号

$\sqrt{5}, \sqrt[3]{6}, \sqrt[4]{7}$

配方法

$\frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$

$y = f (x) = a x^{2} + b x + c$
$= a [x^{2} + \frac{b}{a} x + \frac{c}{a}]$
$= a [x^{2} + 2 * \frac{b}{2 a} x + (\frac{b}{2 a})^{2} - (\frac{b}{2 a})^{2} + \frac{c}{a}]$
$= a [(x + \frac{b}{2 a})^{2} - \frac{b^{2}}{4 a^{2}} + \frac{4 a c}{4 a^{2}}]$
$= a [(x + \frac{b}{2 a})^{2} - \frac{b^{2} - 4 a c}{4 a^{2}}]$
$= a [(x + \frac{b}{2 a})^{2} - (\sqrt{\frac{b^{2} - 4 a c}{4 a^{2}}})^{2}]$
$= a [(x + \frac{b}{2 a})^{2} - (\frac{\sqrt{b^{2} - 4 a c}}{2 a})^{2}]$
$= a [(x + \frac{b}{2 a} + \frac{\sqrt{b^{2} - 4 a c}}{2 a}) (x + \frac{b}{2 a} - \frac{\sqrt{b^{2} - 4 a c}}{2 a})]$
$= a [x + \frac{b \pm \sqrt{b^{2} - 4 a c}}{2 a}]$
$= a [x - \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}]$

$A (x_{1}, y_{1}), B (x_{2}, y_{2})$
距离 $= \sqrt{(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}}$
中点坐标 $= (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$
$\vec{A B} = (x_{2} - x_{1}, y_{2} - y_{1})$

余弦定理

$△ A B C$ ,角 $A, B, C$ ,对应边 $a, b, c$
$c^{2} = a^{2} + b^{2} - 2 a b \cos C$

重力与轨道

$F = G \frac{M_{1} M_{2}}{r^{2}}$
$a = \frac{v^{2}}{r}$

劳伦兹变换

$m = \frac{m_{0}}{\sqrt{1 - (\frac{v}{c})^{2}}}$
$Δ t_{v} = \frac{Δ t}{\sqrt{1 - (\frac{v}{c})^{2}}}$