Question

вычислить средне арифметическое корней уравнения ​

Answer 1

ОДЗ: 5x² - 20 ≠ 0 и x-2 ≠ 0,

x² - 4 ≠ 0

x ≠ 2 и x≠ -2.

$\frac{2-x}{5x^2 - 20} + \frac{1}{x-2} = 1 - \frac{3}{2-x}$

$\frac{2-x}{5x^2 - 20} + \frac{1}{x-2}  = 1 + \frac{3}{x-2}$

$\frac{2-x}{5x^2 - 20} + \frac{1}{x-2} = \frac{x-2 + 3}{x-2}$

$\frac{2-x}{5x^2 - 20} + \frac{1}{x-2} = \frac{x+1}{x-2}$

$\frac{2-x}{5x^2 - 20} + \frac{1 - (x+1)}{x-2} = 0$

$\frac{2-x}{5x^2 - 20} - \frac{x}{x-2} = 0$

$\frac{2-x - x\cdot 5\cdot (x+2)}{5x^2 - 20} = 0$

$\frac{2 - x - 5x^2 - 10x}{5x^2 - 20} = 0$

$\frac{-5x^2 - 11x + 2}{5x^2 - 20} = 0$

$-5x^2 - 11x + 2 = 0$

$5x^2 + 11x - 2 = 0$

$D = 11^2 - 4\cdot (-2)\cdot 5 = 121 + 40 = 161$

$x = \frac{-11 \pm\sqrt{161}}{10}$

$x_1 = \frac{-11 - \sqrt{161}}{10}$

$x_2 = \frac{-11 + \sqrt{161}}{10}$

$\frac{x_1 + x_2}{2} = \frac{1}{2}\cdot \frac{-11 - \sqrt{161} - 11 + \sqrt{161}}{10} =$

$= -\frac{11}{10}  = -1{,}1$