a, \(\frac{2}{5}+\frac{1}{4}\times x=\frac{3}{10}\)
\(\Leftrightarrow\frac{1}{4}\times x=\frac{3}{10}-\frac{2}{5}\)
\(\Leftrightarrow\frac{1}{4}\times x=\frac{-1}{10}\)
\(\Leftrightarrow x=\frac{-1}{10}\div\frac{1}{4}\)
\(\Leftrightarrow x=\frac{-2}{5}\)
Vậy \(x=\frac{-2}{5}\)
b, \(\frac{2}{3}+\frac{2}{3}\div x=\frac{4}{15}\)
\(\Leftrightarrow\frac{2}{3}\div x=\frac{4}{15}-\frac{2}{3}\)
\(\Leftrightarrow\frac{2}{3}\div x=\frac{4}{15}-\frac{2}{3}\)
\(\Leftrightarrow\frac{2}{3}\div x=\frac{-2}{5}\)
\(\Leftrightarrow x=\frac{2}{3}\div\frac{-2}{5}\)
\(\Leftrightarrow\frac{-5}{3}\)
Vậy \(x=\frac{-5}{3}\)
c, \(2\times\left|\frac{2}{3}-x\right|+\frac{1}{4}=\frac{3}{4}\)
\(\Leftrightarrow2\times\left|\frac{2}{3}-x\right|=\frac{3}{4}-\frac{1}{4}\)
\(\Leftrightarrow2\times\left|\frac{2}{3}-x\right|=\frac{1}{2}\)
\(\Leftrightarrow\left|\frac{2}{3}-x\right|=\frac{1}{2}\div2\)
\(\Leftrightarrow\left|\frac{2}{3}-x\right|=\frac{1}{4}\)
\(\Leftrightarrow\hept{\begin{cases}\frac{2}{3}-x=\frac{1}{4}\\\frac{2}{3}-x=\frac{-1}{4}\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=\frac{5}{12}\\x=\frac{11}{12}\end{cases}}\)
Vậy \(x\in\left\{\frac{5}{12};\frac{11}{12}\right\}\)
d, \(3\times\left|\frac{5}{4}-x\right|-\frac{1}{8}=\frac{1}{4}\)
\(\Leftrightarrow3\times\left|\frac{5}{4}-x\right|=\frac{1}{4}+\frac{1}{8}\)
\(\Leftrightarrow3\times\left|\frac{5}{4}-x\right|=\frac{3}{8}\)
\(\Leftrightarrow\left|\frac{5}{4}-x\right|=\frac{3}{8}\div3\)
\(\Leftrightarrow\left|\frac{5}{4}-x\right|=\frac{1}{8}\)
\(\Leftrightarrow\hept{\begin{cases}\frac{5}{4}-x=\frac{1}{8}\\\frac{5}{4}-x=\frac{-1}{8}\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=\frac{9}{8}\\x=\frac{11}{8}\end{cases}}\)
Vậy \(x\in\left\{\frac{9}{8};\frac{11}{8}\right\}\)
