a) \(\left|2x-5\right|=4\)
TH1: \(2x-5=4\Rightarrow2x=9\Rightarrow x=\dfrac{9}{2}\)
TH2: \(2x-5=-4\Rightarrow2x=1\Rightarrow x=\dfrac{1}{2}\)
Vậy \(x=\left\{\dfrac{9}{2};\dfrac{1}{2}\right\}\).
b) \(\dfrac{1}{3}-\left|\dfrac{5}{4}-2x\right|=\dfrac{1}{4}\)
\(\left|\dfrac{5}{4}-2x\right|=\dfrac{1}{3}-\dfrac{1}{4}\)
\(\left|\dfrac{5}{4}-2x\right|=\dfrac{1}{12}\)
TH1: \(\dfrac{5}{4}-2x=\dfrac{1}{12}\Rightarrow2x=\dfrac{7}{6}\Rightarrow x=\dfrac{7}{12}\)
TH2: \(\dfrac{5}{4}-2x=-\dfrac{1}{12}\Rightarrow2x=\dfrac{4}{3}\Rightarrow x=\dfrac{2}{3}\)
A . X = \(\dfrac{1}{2}\)
X = \(4\dfrac{1}{2}\)
B. X = \(\dfrac{7}{12}\)
X = \(\dfrac{2}{3}\)