Question

Нужна помощь)
$\frac{4}{ x^{2} -9} - \frac{x+1}{x-3} =1$

Answer 1

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

Ответ: $\sqrt{6} -1;-\sqrt{6}-1$