\(\left|x-1\right|-3x-2=0\)
\(\Leftrightarrow\left|x-1\right|=2+3x\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=2+3x\\x-1=-2+3x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3x=2+1\\x-3x=\left(-2\right)+1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x=3\\-2x=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy .....
Lại nhỉ hình như vẫn sai p ko???
\(\left|x-1\right|-3x-2=0\)
\(\Rightarrow\left|x-1\right|=3x+2\)
Điều kiện :\(3x-2\ge0\Rightarrow3x\ge-2\Rightarrow x\ge\dfrac{-2}{3}\)
\(\Rightarrow\left[{}\begin{matrix}x-1=3x+2\\x-1=-3x-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-3x=2+1\\x+3x=-2+1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}-2x=3\\4x=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-2}{3}\\x=\dfrac{-1}{4}\end{matrix}\right.\)
vì cả 2 trường hợp trên đều thỏa mãn điều kiện \(x\ge\dfrac{-2}{3}\)
Vậy \(\left[{}\begin{matrix}x=\dfrac{-2}{3}\\x=\dfrac{-1}{4}\end{matrix}\right.\)
ko chép mạng 100% tick cho mink nhé
Ta có :
\(\left|x-1\right|-3x-2=0\)
\(\Leftrightarrow\left|x-1\right|=2+3x\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=2+3x\\x-1=-2+3x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-x=\left(-1\right)-2\\3x-x=\left(-1\right)+2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=-3\\2x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy .....
