\(-2x-\frac{3}{4}=x-\frac{3}{5}\)
\(3x=\frac{3}{5}-\frac{3}{4}=-\frac{3}{20}\)
\(x=-\frac{1}{20}\)
b) \(\left|\frac{x}{2}-\frac{1}{3}\right|=\frac{1}{4}+\frac{3}{2}=\frac{7}{4}\)
=> \(\orbr{\begin{cases}\frac{x}{2}-\frac{1}{3}=\frac{7}{4}\\\frac{x}{2}-\frac{1}{3}=-\frac{7}{4}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{25}{6}\\x=-\frac{17}{6}\end{cases}}}\)
a, \(-2x-\frac{3}{4}=x-\frac{3}{5}\)
\(-2x-\frac{3}{4}-x+\frac{3}{5}=0\)
\(-3x-\frac{3}{20}=0\)
\(\frac{3}{20}=-3x\Leftrightarrow x=\frac{1}{20}\)
\(a,-2x+\frac{-3}{4}=x-\frac{3}{5}\)
\(-2x-x=\frac{3}{4}-\frac{3}{5}\)
\(-3x=\frac{15}{20}-\frac{12}{20}\)
\(-3x=\frac{3}{20}\)
\(x=\frac{3}{20}\div\left(-3\right)\)
\(x=\frac{3}{20}.\frac{-1}{3}\)
\(x=-\frac{1}{20}\)
Vậy \(x=-\frac{1}{20}\).
\(b,\left|\frac{1}{2}x-\frac{1}{3}\right|-\frac{3}{2}=\frac{1}{4}\)
\(\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{1}{4}+\frac{3}{2}\)
\(\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{1}{4}+\frac{6}{4}\)
\(\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{7}{4}\)
\(\Rightarrow\orbr{\begin{cases}\frac{1}{2}x-\frac{1}{3}=\frac{7}{4}\\\frac{1}{2}x-\frac{1}{3}=-\frac{7}{4}\end{cases}}\Rightarrow\orbr{\begin{cases}\frac{1}{2}x=\frac{25}{12}\\\frac{1}{2}x=-\frac{17}{12}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{25}{6}\\x=-\frac{17}{6}\end{cases}}\)
Vậy \(x\in\left\{\frac{25}{6};-\frac{17}{6}\right\}\).
b, \(\left|\frac{1}{2}x-\frac{1}{3}\right|-\frac{3}{2}=\frac{1}{4}\)
\(\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{7}{4}\)
TH1 : \(\frac{1}{2}x-\frac{1}{3}=\frac{7}{4}\Leftrightarrow\frac{1}{2}x=\frac{25}{12}\Leftrightarrow x=\frac{25}{6}\)
TH2 : \(\frac{1}{2}x-\frac{1}{3}=-\frac{7}{4}\Leftrightarrow\frac{1}{2}x=-\frac{17}{12}\Leftrightarrow x=-\frac{17}{6}\)
