Question

Решите уравнение. Методом интервалов
$frac{ x^{2}-5x-6}{ x^{2} -1} leq frac{x-9}{x-1} + frac{2}{x-3}$

Answer 1

$dfrac{x^2-5x-6}{x^2-1} leq dfrac{x-9}{x-1}+ dfrac{2}{x-3} \ dfrac{(x-3)(x^2-5x-6)-(x+1)(x-3)(x-9)-2(x+1)(x-1)}{(x+1)(x-1)(x-3)} leq 0 \ dfrac{x^3-5x^2-6x-3x^2+15x+18-x^3+11x^2-15x-27-2x^2+2}{(x+1)(x-1)(x-3)} leq 0 \ dfrac{x^2-6x-7}{(x-1)(x+1)(x-3)} leq 0 \ \ x^2-6x-7=0 \ x_1+x_2=6 \ x_1x_2=-7 \ x_1=-1 \ x_2=7 \ \ dfrac{(x-7)(x+1)}{(x-1)(x+1)(x-3)} leq 0$

_______-______(-1)_____-____(1)_____+____(3)_____-____(7)

x∈(-∞;-1)U(-1;1)U(3;7]