a) \(\left|8-2x\right|+\left|x-5\right|=7 \)
\(\Rightarrow8-2x+x-5=7\) hoặc \(8-2x+x-5=-7\)
\(\left(1\right)\) \(8-2x+x-5=7\)
\(8-x-5=7\\ 3-x=7\\ x=3-7\\ x=-4\)
\(\left(2\right)\) \(8-2x+x-5=-7\)
\(8-x-5=-7\\ 3-x=-7\\ x=3-\left(-7\right)\\ x=10\)
Vậy \(x=-4\) hoặc \(x=10\)
b) \(\left(\dfrac{2}{3}-x\right)\left(\dfrac{1}{2}-2x\right)=0\)
\(\Rightarrow\left(\dfrac{2}{3}-x\right)=0\) hoặc \(\left(\dfrac{1}{2}-2x\right)=0\)
\(\left(1\right)\left(\dfrac{2}{3}-x\right)=0\\ \Rightarrow x=\dfrac{2}{3}-0\\ -x=\dfrac{2}{3}\\ \Rightarrow x=-\dfrac{2}{3}\)
\(\left(2\right)\left(\dfrac{1}{2}-2x\right)=0\\ \Rightarrow-2x=\dfrac{1}{2}-0\\ -2x=\dfrac{1}{2}\\ x=\dfrac{1}{2}:-2\\ x=-\dfrac{1}{4}\)
Vậy: \(x=-\dfrac{2}{3}\) hoặc \(x=-\dfrac{1}{4}\)
