Question

x-3/x-2+x-7/2-x=20/x2-4

Answer 1

На фотографии подробно.

Answer 2

$\displaystyle \tt \frac{x-3}{x-2}+\frac{x-7}{2-x}=\frac{20}{x^2-4}\: \: \: \: | \: x\ne2, \: x\ne-2\\\displaystyle \tt \frac{x-3}{x-2}+\frac{x-7}{2-x}-\frac{20}{x^2-4}=0\\\displaystyle \tt \frac{x-3}{x-2}+\frac{x-7}{-(x-2)}-\frac{20}{(x-2)(x+2)}=0\\\displaystyle \tt \frac{x-3}{x-2}-\frac{x-7}{x-2}-\frac{20}{(x-2)(x+2)}=0\\\displaystyle \tt \frac{(x+2)(x-3)-(x+2)(x-7)-20}{(x-2)(x+2)}=0\\\displaystyle \tt \frac{x^2-3x+2x-6-(x^2-7x+2x-14)-20}{(x-2)(x+2)}=0$

$\displaystyle \tt \frac{x^2-3x+2x-6-(x^2-5x-14)-20}{(x-2)(x+2)}=0\\\displaystyle \tt \frac{x^2-3x+2x-6-x^2+5x+14-20}{(x-2)(x+2)}=0\\\displaystyle \tt \frac{4x-12}{(x-2)(x+2)}=0\\\displaystyle \tt 4x-12=0\\\displaystyle \tt 4x=12\\\displaystyle \tt x=\frac{12}{4}\\\displaystyle \tt \boxed{\bold{x=3}}$