Lời giải:
ĐKXĐ: $x\neq 0; -1$
PT \(\Leftrightarrow \frac{1}{x^2}-\frac{1}{(x+1)^2}+\left(\frac{3}{x+1}-\frac{1}{(x+1)^2}-2\right)=0\)
\(\Leftrightarrow \frac{(x+1)^2-x^2}{x^2(x+1)^2}+\left[\frac{1}{x+1}-\frac{1}{(x+1)^2}+\frac{2}{x+1}-2\right]=0\)
\(\Leftrightarrow \frac{2x+1}{x^2(x+1)^2}+\left(\frac{1}{x+1}-2\right)\left(1-\frac{1}{x+1}\right)=0\)
\(\Leftrightarrow \frac{2x+1}{x^2(x+1)^2}-\frac{x(2x+1)}{(x+1)^2}=0\)
\(\Rightarrow 2x+1-x^3(2x+1)=0\)
\(\Leftrightarrow (2x+1)(1-x^3)=0\Leftrightarrow (2x+1)(1-x)(x^2+x+1)=0\)
Dễ thấy $x^2+x+1\neq 0$ nên $(2x+1)(1-x)=0$
$\Rightarrow x=-\frac{1]{2}$ hoặc $x=1$ (t/m)
\(\frac{1}{x^2}+\frac{3}{x+1}-\frac{2}{\left(x+1\right)^2}=2\)
\(\Leftrightarrow\frac{\left(x+1\right)^2}{x^2\left(x+1\right)^2}+\frac{3x^2\left(x+1\right)}{x^2\left(x+1\right)^2}-\frac{2x^2}{x^2\left(x+1\right)^2}=\frac{2x^2\left(x+1\right)^2}{x^2\left(x+1\right)^2}\)
\(\Leftrightarrow\left(x+1\right)^2+3x^2\left(x+1\right)-2x^2=2x^2\left(x+1\right)^2\)
\(\Leftrightarrow2x^2+2x+1+3x^3=2x^4+4x^3+2x^2\)
\(\Leftrightarrow2x+1-2x^4-x^3=0\)
\(\Leftrightarrow\left(-2x^3-3x^2-3x-1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x^3-3x^2-3x-1=0\\x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{1}{2}\\x=1\end{matrix}\right.\)
Vậy ..........
