\(\Leftrightarrow5\left(x-1\right)^2=\left(2x+1\right)\left(x+1\right)\)
\(\Leftrightarrow5\left(x^2-2x+1\right)=2x^2+2x+x+1\)
\(\Leftrightarrow5x^2-10x+5-\left(2x^2+3x+1\right)=0\)
\(\Leftrightarrow3x^2-13x+4=0\)
\(\Leftrightarrow x\left(3x-1\right)-4\left(3x-1\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(3x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=4\\x=\frac{1}{3}\end{matrix}\right.\)
ĐKXĐ: \(x\ne1;x\ne-1\)
Ta có: \(\frac{2x+1}{x-1}=\frac{5\left(x-1\right)}{x+1}\)
\(\Leftrightarrow\frac{2x+1}{x-1}-\frac{5\left(x-1\right)}{x+1}=0\)
\(\Leftrightarrow\frac{\left(2x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{5\left(x-1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=0\)
\(\Leftrightarrow\left(2x+1\right)\left(x+1\right)-5\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow2x^2+2x+x+1-5x^2+10x-5=0\)
\(\Leftrightarrow-3x^2+13x-4=0\)
\(\Leftrightarrow-3x^2+x+12x-4=0\)
\(\Leftrightarrow x\left(-3x+1\right)+4\left(3x-1\right)=0\)
\(\Leftrightarrow x\left(1-3x\right)-4\left(1-3x\right)=0\)
\(\Leftrightarrow\left(1-3x\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}1-3x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=1\\x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\x=4\end{matrix}\right.\)(tm đkxđ)
Vậy: \(x\in\left\{\frac{1}{3};4\right\}\)
ĐKXĐ: \(x\ne1,x\ne-1\)
\(\frac{2x+1}{x-1}=\frac{5\left(x-1\right)}{\left(x+1\right)}\Leftrightarrow\left(2x+1\right)\left(x+1\right)=5\left(x-1\right)^2\)
\(\Leftrightarrow5x^2-2x^2-10x-3x+5-1=0\Leftrightarrow3x^2-13x+4=0\Leftrightarrow\left(3x-1\right)\left(x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\x=4\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=\frac{1}{3}\\x=4\end{matrix}\right.\)
