a. \(\left|2-x\right|+\dfrac{3}{4}=6,75\Leftrightarrow\left|2-x\right|=6,75-\dfrac{3}{4}=6\Leftrightarrow\left[{}\begin{matrix}2-x=6\\2-x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=8\end{matrix}\right.\)
Vậy x=-4 hoặc x=8
d. \(\left|3x-1\right|=\left|5-2x\right|\Rightarrow\left[{}\begin{matrix}3x-1=5-2x\\3x-1=2x-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x+2x=5+1\\3x-2x=-5+1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}5x=6\\x=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1,2\\x=-4\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=1,2\\x=-4\end{matrix}\right.\)
c. \(\left|3x-1\right|=1-3x\Rightarrow\left[{}\begin{matrix}3x-1=1-3x;1-3x\ge0\\3x-1=3x-1;3x-1< 0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x+3x=1+1;3x\ge1\\3x-3x=-1+1;3x< 1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}6x=2;x\ge\dfrac{1}{3}\\x\in R;x< \dfrac{1}{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3};x\ge\dfrac{1}{3}\\x\in R,x< \dfrac{1}{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x\in R,x< \dfrac{1}{3}\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=\dfrac{1}{3}\\x\in R,x< \dfrac{1}{3}\end{matrix}\right.\)
