Bài 1: Tìm x
a/ \(2\left|\frac{2}{3}-x\right|+\frac{1}{4}=\frac{3}{4}\)
\(2\left|\frac{2}{3}-x\right|=\frac{3}{4}-\frac{1}{4}\)
\(2\left|\frac{2}{3}-x\right|=\frac{1}{2}\)
\(\left|\frac{2}{3}-x\right|=\frac{1}{2}:2\)
\(\left|\frac{2}{3}-x\right|=\frac{1}{4}\)
\(\Rightarrow\frac{2}{3}-x=\frac{1}{4}\) hoặc \(\frac{2}{3}-x=\frac{-1}{4}\)
TH1: \(\frac{2}{3}-x=\frac{1}{4}\)
\(x=\frac{2}{3}-\frac{1}{4}\)
\(x=\frac{5}{12}\)
TH2: \(\frac{2}{3}-x=\frac{-1}{4}\)
\(x=\frac{2}{3}+\frac{1}{4}\)
\(x=\frac{11}{12}\)
Vậy \(x=\frac{5}{12}\) hoặc \(x=\frac{11}{12}\)
b/ \(\left(2x-1\right)^3=-8\)
\(\left(2x-1\right)^3=\left(-2\right)^3\)
\(\Rightarrow2x-1=-2\)
\(2x\) \(=-2+1\)
\(2x\) \(=-1\)
\(x\) \(=-1:2\)
\(x\) \(=\frac{-1}{2}\)
c/ \(2.\left|x-\frac{5}{7}\right|-\frac{2}{7}=\frac{1}{7}\)
\(2.\left|x-\frac{5}{7}\right|=\frac{1}{7}+\frac{2}{7}\)
\(2.\left|x-\frac{5}{7}\right|=\frac{3}{7}\)
\(\left|x-\frac{5}{7}\right|=\frac{3}{7}:2\)
\(\left|x-\frac{5}{7}\right|=\frac{3}{14}\)
\(\Rightarrow x-\frac{5}{7}=\frac{3}{14}\) hoặc \(x-\frac{5}{7}=\frac{-3}{14}\)
TH1: \(x-\frac{5}{7}=\frac{3}{14}\)
\(x\) \(=\frac{3}{14}+\frac{5}{7}\)
\(x\) \(=\frac{13}{14}\)
TH2: \(x-\frac{5}{7}=\frac{-3}{14}\)
\(x\) \(=\frac{-3}{14}+\frac{5}{7}\)
\(x\) \(=\frac{1}{2}\)
Vậy \(x=\frac{13}{14}\) hoặc \(x=\frac{1}{2}\)