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