Question

СРОЧНО!!!!!!!!! АЛГЕБРА

Answer 1

Ответ:

а) x = $-\frac{5}{3}$ или $-1 \frac{2}{3}$

б) x₁ = $2-\sqrt{7}$ ; x₂ = $2+\sqrt{7}$

Объяснение:

а)

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

$\frac{1}{x+2} - \frac{3}{x+1} = \frac{x}{x^2+3x+2}$ $, x \neq -2, x \neq -1$

$\frac{1}{x+2} - \frac{3}{x+1} - \frac{x}{x^2+3x+2} = 0$

$\frac{1}{x+2} - \frac{3}{x+1} - \frac{x}{x^2+2x+x+2} = 0$

$\frac{1}{x+2} - \frac{3}{x+1} - \frac{x}{x(x+2)+x+2} = 0$

$\frac{1}{x+2} - \frac{3}{x+1} - \frac{x}{(x+2)(x+1)} = 0$

$\frac{x+1-3(x+2)-x}{(x+2)(x+1)} = 0$

$\frac{1-3x-6}{(x+2)(x+1)} = 0$

$\frac{-5-3x}{(x+2)(x+1)} = 0$

$-5-3x = 0$

$-3x=5$

$x = -\frac{5}{3}$ $, x\neq -2, x \neq -1$

$x = -\frac{5}{3}$ или $-1\frac{2}{3}$

б)

$\frac{x}{x+3} - \frac{x-4}{3-x} = 1$

$\frac{x}{x+3} - \frac{x-4}{3-x} = 1$ $, x \neq -3, x\neq 3$

$\frac{x}{x+3} - \frac{x-4}{3-x} - 1 = 0$

$\frac{x(3-x)-(x+3)(x-4)-(x+3)(3-x)}{(x+3)(3-x)} = 0$

$\frac{3x-x^2-(x^2-4x+3x-12)-(3+x)(3-x)}{(x+3)(3-x)} = 0$

$\frac{3x-x^2-(x^2-x-12)-(9-x^2)}{(x+3)(3-x)} = 0$

$\frac{3x-x^2-(x^2-x-12)-9+x^2}{(x+3)(3-x)} = 0$

$\frac{3x-x^2+x+12-9}{(x+3)(3-x)} = 0$

$\frac{4x-x^2+3}{(x+3)(3-x)} = 0$

$4x-x^2+3=0$

$-x^2+4x+3=0$

$x^2-4x-3=0$

$x = \frac{-(-4)+-\sqrt{(-4)^2-4*1*(-3)} }{2*1}$

$x = \frac{4+-\sqrt{16+12} }{2}$

$x = \frac{4+-2\sqrt{7}}{2}$

$x=\frac{4+2\sqrt{7} }{2}$

$x = \frac{4-2\sqrt{7} }{2}$

$x = 2+\sqrt{7} , x \neq -3, x\neq 3$

$x = 2-\sqrt{7} , x \neq -3, x \neq 3$

$x = 2+\sqrt{7}$

$x = 2-\sqrt{7}$

$x$₁ $= 2-\sqrt{7}$ ; $x$₂ $= 2+ \sqrt{7}$