\(\dfrac{2x}{x-1}+\dfrac{3x-2}{x+2}=\dfrac{6}{\left(x-1\right)\left(x+2\right)}\left(ĐKXĐ:x\ne1;x\ne-2\right)\)
\(\Leftrightarrow\dfrac{2x\left(x+2\right)+\left(3x-2\right)\left(x-1\right)}{\left(x-1\right)\left(x+2\right)}=\dfrac{6}{\left(x-1\right)\left(x+2\right)}\)
\(\Rightarrow2x\left(x+2\right)+\left(3x-2\right)\left(x-1\right)=6\)
\(\Leftrightarrow2x^2+4x+3x^2-3x-2x+2-6=0\)
\(\Leftrightarrow5x^2-x-4=0\)
\(\Leftrightarrow5x^2-5x+4x-4=0\)
\(\Leftrightarrow5x\left(x-1\right)+4\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(5x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\5x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(l\right)\\x=\dfrac{-4}{5}\left(n\right)\end{matrix}\right.\)
Vậy .....................
ĐK: $x\ne 1; x\ne -2$.
\(\dfrac{2x}{x-1}+\dfrac{3x-2}{x+2}=\dfrac{6}{\left(x-1\right)\left(x+2\right)}\\ \Rightarrow2x\left(x+2\right)+\left(3x-2\right)\left(x-1\right)=6\\ \Leftrightarrow2x^2+4x+3x^2-3x-2x+2=6\\ \Leftrightarrow5x^2-x-4=0\\ \Leftrightarrow5x^2-5x+4x-4=0\\ \Leftrightarrow5x\left(x-1\right)+4\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(5x+4\right)=0\)
$\Leftrightarrow x-1=0$ hoặc $5x+4=0$
$\Leftrightarrow x=1$ (loại) hoặc $x=\dfrac{-4}5$
