Điều kiện : \(x\ne\pm1\)
\(\frac{x+4}{x+1}+\frac{x}{x-1}=\frac{2x^2}{x^2-1}\)
\(\Rightarrow\frac{\left(x+4\right)\left(x-1\right)+x\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}=\frac{2x^2}{\left(x+1\right)\left(x-1\right)}\)
\(\Rightarrow\left(x+4\right)\left(x-1\right)+x\left(x+1\right)=2x^2\)
\(\Rightarrow x^2-x+4x-4+x^2+x=2x^2\)
\(\Rightarrow2x^2+4x+4=2x^2\)
\(\Rightarrow\left(x^2+4x+4\right)=2x^2-x^2\)
\(\Rightarrow\left(x+2\right)^2=x^2\)
\(\Rightarrow\left|x+2\right|=\left|x\right|\)
\(\Rightarrow\left[\begin{array}{nghiempt}x+2=x\\x+2=-x\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x\in\varnothing\\x=1\end{array}\right.\) (loại )
Vậy phương trình vô nghiệm
