用户:R96340/LaTeX练习

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

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

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

1

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

$= a [(x + \frac{b}{2 a})^{2} - \frac{b^{2}}{4 a^{2}} + \frac{4 a c}{4 a^{2}}] = a [(x + \frac{b}{2 a})^{2} - \frac{b^{2} - 4 a c}{4 a^{2}}]$

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

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

$= a [x - \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}]$

2

$A (x, y), B (x, y)$

距离 $= \sqrt{(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}}$

中点坐标 $= (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$

$\vec{A B} = (x_{2} - x_{1}, y_{2} - y_{1})$

3

$△ A B C$ 角 $∠ A, B, C$ 对应边 $\overline{a}, \overline{b}, \overline{c}$

$C^{2} = a^{2} + b^{2} - 2 a b c o s C$

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

$a = \frac{V^{2}}{r}$

4

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

$△ t_{v} = \frac{△ t}{\sqrt{1 - (\frac{v}{c})^{2}}}$