Question

Помогите решить уравнение 1/(x^2+2x)-1/(x+1)^2=1/30

Answer 1

ОДЗ:

$\displaystyle\begin{cases}x^2+2x\ne0\\x+1\ne0\end{cases}\to\begin{cases}x\ne-2\\x\ne-1\\x\ne0\end{cases}$

$\displaystyle\frac{1}{x^2+2x}-\frac{1}{(x+1)^2}=\frac{1}{30}|*30(x^2+2x)(x+1)^2\\30(x+1)^2-30(x^2+2x)=(x^2+2x)(x+1)^2\\30(x^2+2x+1)-30(x^2+2x)=(x^2+2x)(x^2+2x+1)\\\\x^2+2x=a\\30=a(a+1)\\a^2+a-30=0\\a_{1,2}=\frac{-1\pm\sqrt{1+120}}{2}=\frac{-1\pm11}{2}\\a_1=-6\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ a_2=5\\x^2+2x+6=0\ \ \ \ \ \ \ x^2+2x-5=0\\x_{1,2}=-1\pm\sqrt{1-6}\ \ x_{1,2}=-1\pm\sqrt{1+5}\\x\in\varnothing\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x_1=-1+\sqrt6;x_2=-1-\sqrt6$