Question

Решите уравнение:
$\frac{4}{ x^{2} -9} - \frac{x+1}{x-3} =1$

Answer 1

$\frac{4}{(x-3)(x+3)}- \frac{x+1}{x-3} -1=0 \\ \\ x \neq \pm3 \\ \\ \frac{4-(x+1)(x+3)-(x^2-9)}{(x-3)(x+3)}=0 \\ \\ 4-(x^2+3x+x+3)-x^2+9=0 \\ 4-x^2-4x-3-x^2+9=0 \\ 2x^2+4x-10=0 \\ x^2+2x-5=0 \\ D=4+20=24=(2 \sqrt{6})^2 \\ \\ x_1= \frac{-2-2 \sqrt{6} }{2} \ (:2)=-1- \sqrt{6} \\ x_2=-1+ \sqrt{6}$