\(\dfrac{5}{3x+2}=2x-1\)
\(\Leftrightarrow\left(2x-1\right)\left(3x+2\right)=5\)
\(\Leftrightarrow6x^2-x-2=5\)
\(\Leftrightarrow6x^2-x-7=0\)
\(\Leftrightarrow\left(6x^2+6x\right)-\left(7x+7\right)=0\)
\(\Leftrightarrow6x\left(x+1\right)-7\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(6x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\6x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{7}{6}\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=-1\\x=\dfrac{7}{6}\end{matrix}\right.\)
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