\(\left|x+1\right|-4x=0\\ \Leftrightarrow\left|x+1\right|=4x\)
Ta có : \(\left\{{}\begin{matrix}x+1\ge0\Leftrightarrow x\ge-1\\x+1< 0\Leftrightarrow x< -1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=4x\\\left(-x+1\right)=4x\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x-4x=-1\\-x-1=4x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-3x=-1\\-x-4x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\left(tm\right)\\x=-\dfrac{1}{5}\left(tm\right)\end{matrix}\right.\)
`@ Kidd`
\(\left|x+1\right|-4x=0\)
\(\Leftrightarrow\left|x+1\right|=4x\left(1\right)\)
\(\text{Ta có:}\left\{{}\begin{matrix}x+1\ge0\text{ hay }x\ge-1\\x+1< 0\text{ hay }x< -1\end{matrix}\right.\)
\(TH1:x+1=4x\)
\(\Leftrightarrow x-4x=-1\)
\(\Leftrightarrow-3x=-1\)
\(\Leftrightarrow x=\dfrac{1}{3}\left(\text{nhận}\right)\)
\(TH2:-\left(x+1\right)=4x\)
\(\Leftrightarrow-x-1=4x\)
\(\Leftrightarrow-x-4x=1\)
\(\Leftrightarrow-5x=1\)
\(\Leftrightarrow x=\dfrac{-1}{5}\left(\text{nhận}\right)\)
\(\text{Vậy phương trình (1) có tập nghiệm là }S=\left\{\dfrac{1}{3};\dfrac{-1}{5}\right\}\)
