Question

17 БАЛЛОВ

Answer 1

$\frac{3x-5}{x^2-1} = \frac{3x+2}{x^2+x} - \frac{6x-5}{x^2-x}$
$\frac{3x-5}{(x-1)(x+1)} - \frac{3x+2}{x(x+1)}+ \frac{6x-5}{x(x-1)}=0$
$\frac{x(3x-5)-(3x+2)(x-1)+(6x-5)(x+1)}{x(x-1)(x+1)} =0$
ОДЗ:
$x \neq 0$      
$x \neq 1$
$x \neq -1$
${x(3x-5)-(3x+2)(x-1)+(6x-5)(x+1)}=0$
$3x^2-5x-(3x^2-3x+2x-2)+6x^2+6x-5x-5}=0$
$3x^2-5x-(3x^2-x-2)+6x^2+x-5}=0$
$3x^2-5x-3x^2+x+2+6x^2+x-5}=0$
$6x^2-3x-3=0$
$2x^2-x-1=0$
$D=(-1)^2-4*2*(-1)=9$
$x_1= \frac{1+3}{4}=1$  ∅
$x_2= \frac{1-3}{4} =-0.5$

Ответ: -0.5