ĐKXĐ: \(x\ne\left\{-1;0\right\}\)
\(\Leftrightarrow2x^2+3\left(x+1\right)\left(x-1\right)=5x\left(x+1\right)\)
\(\Leftrightarrow2x^2+3x^2-3=5x^2+5x\)
\(\Leftrightarrow5x=-3\Rightarrow x=-\frac{3}{5}\)
Ta có: \(\frac{2x}{x+1}+\frac{3\left(x-1\right)}{x}=5\)
⇔\(\frac{2x^2}{x\left(x+1\right)}+\frac{3\left(x-1\right)\left(x+1\right)}{x\left(x+1\right)}-\frac{5x\left(x+1\right)}{x\left(x+1\right)}=0\)
⇔\(2x^2+3x^2-3-5x\left(x+1\right)=0\)
\(\Leftrightarrow5x^2-3-5x^2-5x=0\)
\(\Leftrightarrow-3-5x=0\)
\(\Leftrightarrow-\left(3+5x\right)=0\)
\(\Leftrightarrow3+5x=0\)
\(\Leftrightarrow5x=-3\)
\(\Leftrightarrow x=\frac{-3}{5}\)
Vậy: \(x=\frac{-3}{5}\)
