(16/x^2+x)-(6/x^2-x)=1/x решите пожалуйста))

$\frac{16}{ x^{2} +x} - \frac{6}{ x^{2} -x} = \frac{1}{x} \\ \frac{16}{ x(x+1)} - \frac{6}{ x(x-1)} = \frac{1}{x} /*x(x+1)(x-1) \neq 0\\ x(x+1)(x-1) \neq 0 \\ x \neq 0 \\ x \neq -1\\x \neq 1 \\ \frac{16x(x+1)(x-1)}{x(x+1)} - \frac{6x(x+1)(x-1)}{x(x-1)} = \frac{x(x+1)(x-1)}{x} \\ 16x-16-6x-6- x^{2} +1=0\\- x^{2} +10x-21=0/*(-1)\\ x^{2} -10x+21=0 \\ D=100-84=16 \\ \sqrt{D} =4 \\ x_{1} = \frac{10+4}{2} =7 \\$
$x_{2} = \frac{10-4}{2} =3$