Question

решите пж............​

Answer 1

Ответ:   №4 .

$\displaystyle \dfrac{x-3}{x-1}=\frac{x+6}{x+2}+\frac{x-6}{x-2}-\frac{x+3}{x+1}\\\\\\ \dfrac{x-3}{x-1}+\frac{x+3}{x+1}=\frac{x+6}{x+2}+\frac{x-6}{x-2}\\\\\\\frac{(x-3)(x+1)+(x+3)(x-1)}{(x-1)(x+1)}=\frac{(x+6)(x-2)+(x-6)(x+2)}{(x-2)(x+2)}\\\\\\\frac{x^2-2x-3+x^2+2x-3}{x^2-1}=\frac{x^2+4x-12+x^2-4x-12}{x^2-4}\\\\\\\frac{2x^2-6}{x^2-1}=\frac{2x^2-24}{x^2-4}\\\\\\\frac{x^2-3}{x^2-1}=\frac{x^2-12}{x^2-4}\\\\\\(x^2-3)(x^2-4)=(x^2-12)(x^2-1)\\\\x^4-7x^2+12=x^4-13x^2+12\\\\13x^2-7x^2=0\ \ ,\ \ 6x^2=0\\\\x^2=0$

$x=0\\\\x+1=0+1=1$