Question

Решение уравнений. Урок 4
Реши уравнение:
ПОМОГИТЕ

Answer 1

$OTBET: \boxed {x=-4}$

Объяснение:

$\displaystyle\\\frac{x^2-12}{x^2-4} +\frac{x}{x-2} =1\\\\\\\frac{x^2-12}{(x-2)(x+2)} +\frac{x}{x-2} =1$

ОДЗ: х-2≠0    х≠2    х+2≠0    х≠-2

$\displaystyle\\\frac{x^2-12+x(x+2)}{(x-2)(x+2)} =1\\\\\\\frac{x^2-12+x^2+2x}{x^2-4} =1\\\\\\\frac{2x^2+2x-12}{x^2-4}-1=0 \\\\\\\frac{2x^2+2x-12-(x^2-4)}{x^2-4}=0\\\\\\\frac{2x^2+2x-12-x^2+4}{x^2-4} =0\\\\\\\frac{x^2+2x-8}{x^2-4} =0 \\\\\\x^2-4\neq 0\ \ \ \ \Rightarrow\\\\\\x^2+2x-8=0$

$x^2+4x-2x-8=0\\\\(x^2+4x)-(2x+8)=0\\\\x(x+4)-2(x+4)=0\\\\(x+4)(x-2)=0$

x+4=0

x₁=-4 ∈ ОДЗ

x-2=0

х₂=2 ∉ ОДЗ