Question

Решите уравнение:
$\frac{x+21}{x^{2}-9} - \frac{x}{x+3} = 0$

Answer 1

$\frac{x+21}{(x-3)(x+3)}- \frac{x}{x+3}=0 \\ \\ \frac{x+21-x(x-3)}{(x-3)(x+3)}=0 \\ \frac{x+21-x^2+3x}{(x-3)(x+3)}=0 \\ *x^2-9 \neq 0 =\ \textgreater \ x \neq ^+_--3\\ -x^2+4x+21=0 \\ D=16+4*21=16+84=100 \\ \\ x_1= \frac{-4+10}{2}=-3$ 
$\\ \\ x_2 = \frac{-4-10}{2}=-7$
-3∉