ĐKXĐ:\(x\ne-1;x\ne4\)
\(\frac{2x}{x+1}=\frac{x^2-x+8}{\left(x+1\right)\left(x-4\right)}\)
\(\Leftrightarrow2x=\frac{x^2-x+8}{x-4}\)
\(\Leftrightarrow2x^2-8x=x^2-x+8\)
\(\Leftrightarrow x^2-7x-8=0\)
\(\Leftrightarrow x\left(x+1\right)-8\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=8\end{matrix}\right.\)
Vậy \(S=\left\{-1;8\right\}\)
\(\frac{2x}{x+1}=\frac{x^2-x+8}{\left(x+1\right)\left(x-4\right)}\) (ĐK: \(x\ne-1;x\ne4\))
\(\Leftrightarrow\frac{2x\left(x-4\right)}{\left(x+1\right)\left(x-4\right)}=\frac{x^2-x+8}{\left(x+1\right)\left(x-4\right)}\)
\(\Rightarrow2x^2-8x=x^2-x+8\)
\(\Leftrightarrow2x^2-8x-x^2+x-8=0\)
\(\Leftrightarrow x^2-7x-8=0\)
\(\Leftrightarrow x^2-8x+x-8=0\)
\(\Leftrightarrow\left(x^2-8x\right)+\left(x-8\right)=0\)
\(\Leftrightarrow x\left(x-8\right)+\left(x-8\right)=0\)
\(\Leftrightarrow\left(x-8\right)\left(x-1\right)=0\)
\(\Leftrightarrow x-8=0\) hoặc \(x-1=0\)
* TH1: \(x-8=0\)
\(\Leftrightarrow x=8\) (nhận)
*TH2: \(x-1=0\)
\(\Leftrightarrow x=1\) (nhận)
Vậy pt có tập nghiệm S= \(\left\{8;1\right\}\)
