ĐKXĐ: x≠-1; x≠0
Ta có: \(\frac{2x}{x+1}+\frac{3\left(x-1\right)}{x}=5\)
\(\Leftrightarrow\frac{2x^2}{x\left(x+1\right)}+\frac{3\left(x-1\right)\left(x+1\right)}{x\left(x+1\right)}-\frac{5x\left(x+1\right)}{x\left(x+1\right)}=0\)
\(\Leftrightarrow2x^2+3x^2-3-5x^2-5x=0\)
\(\Leftrightarrow-3-5x=0\)
\(\Leftrightarrow5x=-3\)
hay \(x=\frac{-3}{5}\)(tm)
Vậy: \(x=\frac{-3}{5}\)
\(\frac{2x^2}{x^2+x}+\frac{3x^2-3}{x^2+x}=\frac{5x^2+5x}{x^2+x}\)
\(2x^2+3x^2-3=5x^2+5x\)
\(-5x=-3\)
x=\(\frac{3}{5}\)
