a)\(\frac{9}{4}\cdot\left|x\right|-\frac{5}{2}=\frac{8}{3}\)\(\Rightarrow\frac{9}{4}\cdot\left|x\right|=\frac{8}{3}+\frac{5}{2}\Rightarrow\frac{9}{4}\cdot\left|x\right|=\frac{31}{6}\)
\(\Rightarrow\left|x\right|=\frac{31}{6}:\frac{9}{4}\Rightarrow\left|x\right|=\frac{31}{6}\cdot\frac{4}{9}\Rightarrow\left|x\right|=\frac{62}{27}\)
\(\Rightarrow\hept{\begin{cases}x=\frac{62}{27}\\x=-\frac{62}{27}\end{cases}}\)
b)\(\frac{1}{2}\cdot\left|x\right|+\frac{3}{4}=\frac{2}{3}\Rightarrow\frac{1}{2}\cdot\left|x\right|=\frac{2}{3}-\frac{3}{4}\Rightarrow\frac{1}{2}\cdot\left|x\right|=-\frac{1}{12}\)
\(\Rightarrow\left|x\right|=-\frac{1}{12}:\frac{1}{2}\Rightarrow\left|x\right|=-\frac{1}{12}\cdot2\Rightarrow\left|x\right|=-\frac{1}{6}\)
Ta có\(\left|x\right|\ge0\)mà \(-\frac{1}{6}\le0\)
Do đó ko có giá trị của x thỏa mãn
c)\(\left|x+\frac{1}{5}\right|-4=-2\Rightarrow\left|x+\frac{1}{5}\right|=-2+4\Rightarrow\left|x+\frac{1}{5}\right|=2\)
\(\Rightarrow\hept{\begin{cases}x+\frac{1}{5}=2\Rightarrow x=2-\frac{1}{5}\Rightarrow x=\frac{9}{5}\\x+\frac{1}{5}=-2\Rightarrow x=-2-\frac{1}{5}\Rightarrow x=-\frac{11}{5}\end{cases}}\)
