\(\left(3-\frac{1}{2}x\right)\left(\left|x+\frac{3}{4}\right|-\frac{5}{6}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3-\frac{1}{2}x=0\\\left|x+\frac{3}{4}\right|-\frac{5}{6}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x=3\\\left|x+\frac{3}{4}\right|=\frac{5}{6}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=6\\x=\frac{1}{12}\\x=\frac{-19}{12}\end{cases}}\)
\(\left(3-\frac{1}{2}x\right)\cdot\left(\left|x+\frac{3}{4}\right|-\frac{5}{6}\right)=0\)
\(\Rightarrow\hept{\begin{cases}3-\frac{1}{2}x=0\\\left|x+\frac{3}{4}\right|-\frac{5}{6}=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=6\\x+\frac{3}{4}=\pm\frac{5}{6}\end{cases}}\)
Ta có
\(x+\frac{3}{4}=\pm\frac{5}{6}\)
\(\hept{\begin{cases}x+\frac{3}{4}=\frac{5}{6}\\x+\frac{3}{4}=-\frac{5}{6}\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{1}{12}\\x=-\frac{19}{12}\end{cases}}}\)
Vậy \(x\in\left\{3;\frac{1}{2};-\frac{19}{12}\right\}\)
\(\left(3-\frac{1}{2}.x\right).\left(|x+\frac{3}{4}|-\frac{5}{6}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3-\frac{1}{2}.x=0\\|x+\frac{3}{4}|-\frac{5}{6}=0\end{cases}\Rightarrow\orbr{\begin{cases}\frac{1}{2}.x=3\\|x+\frac{3}{4}|=\frac{5}{6}\end{cases}\Rightarrow}\orbr{\begin{cases}x=6\\x+\frac{3}{4}=\frac{5}{6}\end{cases}\Rightarrow}\orbr{\begin{cases}x=6\\x=\frac{1}{12}\end{cases}}}\)
\(\left(3-\frac{1}{2}x\right).\left(\left|x+\frac{3}{4}\right|-\frac{5}{6}\right)=0\)
=> \(3-\frac{1}{2}x=0\) học \(\left| x+\frac{3}{4}\right|-\frac{5}{6}=0\)
=> \(\frac{1}{2}x=3\) \(\left|x+\frac{3}{4}\right|=\frac{5}{6}\)
= \(x=3:\frac{1}{2}=6\) => + \(x+\frac{3}{4}=\frac{5}{6}\) + \(x+\frac{3}{4}=\frac{-5}{6}\)
\(x=\frac{5}{6}-\frac{3}{4}=\frac{1}{12}\) x = \(\frac{-5}{6}-\frac{3}{4}=\frac{-19}{12}\)
vy x = \(6\) học x = \(\frac{1}{12}\) học x = \(\frac{-19}{12}\)
