a,\(\frac{1}{3}x=-2-\frac{2}{3}=\frac{-8}{3}\)
\(x=\frac{-8}{3}:\frac{1}{3}=\frac{-8}{3}.\frac{3}{1}=-8\)
\(b,\left[x+\frac{1}{1}\right]^2+\frac{5}{6}=\frac{7}{8}\)
\(\Rightarrow\left[x+1\right]^2=\frac{7}{8}-\frac{5}{6}\)
\(\Rightarrow\left[x+1\right]^2=\frac{7\cdot3}{24}-\frac{5\cdot4}{24}\)
\(\Rightarrow\left[x+1\right]^2=\frac{21}{24}-\frac{20}{24}\)
\(\Rightarrow\left[x+1\right]^2=\frac{1}{24}\)
\(\Rightarrow x\in\left\{\varnothing\right\}\)
\(c,\left[3x+\frac{3}{5}\right]\left[\left|x\right|-\frac{1}{4}\right]=0\)
\(\Rightarrow\hept{\begin{cases}3x+\frac{3}{5}=0\\\left|x\right|-\frac{1}{4}=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}3x=0-\frac{3}{5}\\\left|x\right|=0+\frac{1}{4}\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}3x=\frac{-3}{5}\\\left|x\right|=0+\frac{1}{4}\\\left|x\right|=0+\left[\frac{-1}{4}\right]\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=\frac{-1}{5}\\x=\frac{1}{4}\\x=-\frac{1}{4}\end{cases}}\)
Vậy \(x\in\left\{-\frac{1}{5};\pm\frac{1}{4}\right\}\)
