1/ \(\frac{2x+1}{x-1}=\frac{5\left(x-1\right)}{x+1}\)
\(\Leftrightarrow\left(2x+1\right)\left(x+1\right)=5\left(x-1\right)^2\)
\(\Leftrightarrow2x^2+3x+1=5x^2-10x+5\)
\(\Leftrightarrow3x^2-13x+4=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\x=4\end{matrix}\right.\)
KL: Vậy $\orpt{\begin{matrix}x=\frac{1}{3}\\x=4\end{matrix}}$
2/ \(\frac{3x-2}{x+7}=\frac{6x-1}{2x-3}\)
\(\Leftrightarrow\left(3x-2\right)\left(2x-3\right)=\left(6x-1\right)\left(x+7\right)\)
\(\Leftrightarrow6x^2-13x+6=6x^2+41x-7\)
\(\Leftrightarrow54x=13\)
\(\Leftrightarrow x=\frac{13}{54}\)
KL: Vậy .................
Mình bổ sung thêm điều kiện cho câu 1: \(x\ne1,x\ne-1\)
Bổ sung thêm điều kiện cho câu 2: \(x\ne-7,x\ne\frac{3}{2}\)
