đkxđ: x≠\(\pm1\)
pt <=> \(\left(2x+1\right)\left(x+1\right)=5\left(x-1\right)^2\)
\(\Leftrightarrow2x^2+3x+1=5x^2-10x+5\)
\(\Leftrightarrow3x^2-13x+4=0\)
\(\Leftrightarrow3x^2-12x-x+4=0\)
\(\Leftrightarrow3x\left(x-4\right)-\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{1}{3}\end{matrix}\right.\)(TM)
Vậy.............
\(\dfrac{2x+1}{x-1}=\dfrac{5\left(x-1\right)}{x+1}\left(x\ne\pm1\right)\)
\(\Leftrightarrow\dfrac{\left(2x+1\right)\left(x+1\right)-5\left(x-1\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=0\)
\(\Leftrightarrow2x^2+3x+1-5x^2+10x-5=0\)
\(\Leftrightarrow-3x^2+12x+x-4=0\)
\(\Leftrightarrow-3x\left(x-4\right)+\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\left(TMĐKXĐ\right)\\x=\dfrac{1}{3}\left(TMĐKXĐ\right)\end{matrix}\right.\)
KL.........
