\(\left|3x-1\right|=x+2\)
\(\Rightarrow\left[{}\begin{matrix}3x-1=x+2\left(đk:x\ge\dfrac{1}{3}\right)\\-3x+1=x+2\left(đk:x< \dfrac{1}{3}\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}3x=x+3\Rightarrow x=\dfrac{2}{3}\left(TM\right)\\-3x=x-1\Rightarrow x=\dfrac{1}{4}\left(TM\right)\end{matrix}\right.\)
\(\left|x-2\right|+\left|3x-1\right|=0\)
\(\left\{{}\begin{matrix}\left|x-2\right|\ge0\\\left|3x-1\right|\ge0\end{matrix}\right.\)
\(\Rightarrow\left|x-2\right|+\left|3x-1\right|\ge0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left|x-2\right|=0\Rightarrow x=2\\\left|3x-1\right|=0\Rightarrow3x=1\Rightarrow x=\dfrac{1}{3}\end{matrix}\right.\)
Vì \(2\ne\dfrac{1}{3}\Rightarrow x\in\varnothing\)
\(\left|2x-1\right|=\left|x+1\right|\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=x+1\\-2x-1=-x-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=x+2\Rightarrow x=2\\-2x=-x-2\Rightarrow x=2\end{matrix}\right.\)
a) Ta có :
\(\left|3x-1\right|=x+2\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+2=3x-1\\x+2=-\left(3x-1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3+1=3x-x\\x+3x=-2+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=4\\4x=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=\dfrac{-1}{4}\end{matrix}\right.\)
Vậy ......................
b) Ta có :
\(\left|x-2\right|+\left|3x-1\right|=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left|x-2\right|=0\\\left|3x-1\right|=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy ..................
c) Ta có :
\(\left|2x-1\right|=\left|x+1\right|\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-1=x+1\\2x-1=-x-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-x=1+1\\2x+x=-1+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\3x=0\Leftrightarrow x=0\end{matrix}\right.\)
Vậy ..............
C.
\(\left|2x-1\right|=\left|x+1\right|\\ \Leftrightarrow\left(2x-1\right)^2=\left(x+1\right)^2\\ \Leftrightarrow\left(2x-1\right)^2-\left(x+1\right)^2=0\\ \Leftrightarrow\left(2x-1+x+1\right)\left(2x-1-x-1\right)=0\\ \Leftrightarrow3x\left(x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
vậy...
