\(\frac{4}{x+1}=\frac{2}{3x+1}\Leftrightarrow4\left(3x+1\right)=2\left(x+1\right)\Leftrightarrow12x+4=2x+2\)
\(\Leftrightarrow12x-2x=2-4\Leftrightarrow10x=-2\Leftrightarrow\frac{-1}{5}\)
Vậy x=-1/5
\(\frac{4}{x+1}=\frac{2}{3x+1}\left(x\ne-1;x\ne-\frac{1}{3}\right)\)
=> \(4\left(3x+1\right)=2\left(x+1\right)\)
=> \(12x+4=2x+2\)
=> \(12x-2x=2-4\)
=> \(10x=-2\)
=> \(5x=-1\)(chia cho 5)
=> \(x=-\frac{1}{5}\left(tm\right)\)
Vậy \(x=-\frac{1}{5}\)
\(\frac{4}{x+1}=\frac{2}{3x+1}\)
\(\Rightarrow4\left(3x+1\right)=2\left(x+1\right)\)
\(\Leftrightarrow12x+4-2x-2=0\)
\(\Leftrightarrow10x+2=0\)
\(\Leftrightarrow10x=-2\)
\(\Leftrightarrow x=\frac{-1}{5}\)
#H
