A) ĐKXĐ x≠1;x≠-1
\(\dfrac{2x+1}{x-1}=\dfrac{5\left(x-1\right)}{x+1}\)
⇔(2x+1)(x+1)=(x-1)5(x-1)
⇔2x2+2x+x+1=5x2+5-5x-5x
⇔2x2-5x2+2x+x+5x+5x-5+1=0
⇔-3x2+13x-4=0
⇔-(3x2-13x+4)=0
⇔-(3x2-x-12x+4)=0
⇔-[x(3x-1)-4(3x-1)]=0
⇔-(x-4)(3x-1)=0
⇔(4-x)(3x-1)=0
⇔\(\left[{}\begin{matrix}4-x=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=\dfrac{1}{3}\left(tm\right)\end{matrix}\right.\)
vậy S=\(\left\{4;\dfrac{1}{3}\right\}\)
b, ĐKXĐ: \(x\ne2;x\ne4\)
\(\dfrac{x-3}{x-2}+\dfrac{x-2}{x-4}=-1\)
\(\Leftrightarrow\left(x-3\right)\left(x-4\right)+\left(x-2\right)^2+\left(x-2\right)\left(x-4\right)=0\)
\(\Leftrightarrow x^2-7x+12+x^2-4x+4+x^2-6x+8=0\)
\(\Leftrightarrow3x^2-17x+24=0\)
\(\Leftrightarrow3x\left(x-3\right)-8\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(3x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\3x-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{8}{3}\end{matrix}\right.\) (tm)
