ĐKXĐ: \(x\ne0;x\ne-1\)
\(\dfrac{1}{x^2}+\dfrac{3}{x+1}-\dfrac{2}{\left(x+1\right)^2}=2\)
\(\Leftrightarrow\dfrac{\left(x+1\right)^2+3x^2\left(x+1\right)-2x^2}{x^2\left(x+1\right)^2}=2\)
\(\Rightarrow x^2+2x+1+3x^3+3x^2-2x^2=2x^2\left(x+1\right)^2\)
\(\Leftrightarrow3x^3+2x^2+2x+1=2x^4+4x^3+2x^2\)
\(\Leftrightarrow2x^4+x^3-2x-1=0\)
\(\Leftrightarrow x^3\left(2x+1\right)-\left(2x+1\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(x^3-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=0\\x^3-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\left(TM\right)\\x=1\left(TM\right)\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là:
\(S=\left\{-\dfrac{1}{2};1\right\}\)
