\(\dfrac{2}{x+1}=\dfrac{4}{x-5}\\ ĐK:\left\{{}\begin{matrix}x\ne-1\\x\ne5\end{matrix}\right.\\ \Leftrightarrow2.\left(x-5\right)=4.\left(x+1\right)\\ \Leftrightarrow2x-10=4x+4\\ \Leftrightarrow2x-4x=4+10\\ \Leftrightarrow-2x=14\\ \Leftrightarrow x=-7\left(t/m\right)\)
\(\dfrac{2}{x+1}=\dfrac{4}{x-5}\left(x\ne-1;x\ne5\right)\)
suy ra: \(2\left(x-5\right)=4\left(x+1\right)\\ < =>2x-10=4x+4\\ < =>2x-4x=4+10\\< =>-2x=14\\ < =>x=-7\)
\(\dfrac{2}{x+1}=\dfrac{4}{x-5}\\ \Leftrightarrow2.\left(x-5\right)=4.\left(x+1\right)\\ \Leftrightarrow2x-10=4x+4\\ \Leftrightarrow2x-4x=10+4\\ \Leftrightarrow-2x=14\\ \Leftrightarrow x=14:\left(-2\right)\\ \Leftrightarrow x=-7\)
\(\dfrac{2}{x+1}=\dfrac{4}{x-5}\)
ĐKXĐ: \(\left[{}\begin{matrix}x+1\ne0\\x-5\ne0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ne-1\\x\ne5\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{2\left(x-5\right)}{\left(x+1\right)\left(x-5\right)}=\dfrac{4\left(x+1\right)}{\left(x-5\right)\left(x+1\right)}\)
\(\Leftrightarrow2x-10=4x+4\)
\(\Leftrightarrow2x-4x=4+10\)
\(\Leftrightarrow-2x\) \(=14\)
\(\Leftrightarrow x\) \(=\dfrac{14}{-2}=-7\) (nhận)
Vậy S = {-7}
\(\dfrac{2}{x+1}=\dfrac{4}{x-5}\text{ĐKXĐ}:x\ne-1;5\)
\(\Leftrightarrow\dfrac{2\left(x-5\right)}{\left(x+1\right)\left(x-5\right)}=\dfrac{4\left(x+1\right)}{\left(x+1\right)\left(x-5\right)}MTC:\left(x+1\right)\left(x-5\right)\)
\(\Rightarrow2x-10=4x+4\)
\(\Leftrightarrow2x-10-4x-4=0\)
\(\Leftrightarrow-2x-14=0\)
\(\Leftrightarrow-2x=14\)
\(\Leftrightarrow x=-7\left(\text{nhận}\right)\)
\(\text{Vậy phương trình có tập nghiệm là }S=\left\{-7\right\}\)
