Question

$\frac{1}{x(x+2)} - \frac{1}{(x+1)^2} = \frac{1}{12}$

Answer 1

x∉0 x∉-2 x∉-1

$\frac{1}{x(x+2)} - \frac{1}{(x+1)^2} = \frac{1}{12}\\\frac{(x+1)(x+1)-x(x+2)}{x(x+2)(x+1)(x+1)} =\frac{1}{12}\\\frac{x^{2}+2x+1-x^{2}-2x }{x(x+2)(x+1)(x+1)}=\frac{1}{12}\\\frac{1 }{x(x+2)(x+1)(x+1)}=\frac{1}{12}\\\\x(x+2)(x+1)(x+1)=12\\\\(x^{2} +2x)(x^{2} +2x+1)=12$  $t=x^{2} +2x\\t*(t+1)=12\\t^{2} +t -12=0\\D=1+48=49\\t1=(-1+7)/2=3\\\\t2=(-1-7)/2=-4$

$x^{2} +2x-3=0\\x=1\ x=-3\\\\x^{2} +2x+4=0\\D<0$

Ответ: x=1, x=-3