a)\(\dfrac{11}{12}-\left(\dfrac{2}{3}+x\right)=\dfrac{2}{3}\)
\(\dfrac{2}{3}+x=\dfrac{11}{12}-\dfrac{2}{3}\)
\(\dfrac{2}{3}+x=\dfrac{1}{4}\)
\(x=\dfrac{1}{4}-\dfrac{2}{3}\)
\(x=\dfrac{-5}{12}\)
Vậy x=\(\dfrac{-5}{12}\)
b)\(1\dfrac{1}{3}:0,8=\dfrac{2}{3}:0,1.x\)
\(\dfrac{4}{3}:0,8=\dfrac{20}{3}.x\)
\(\dfrac{5}{3}=\dfrac{20}{3}.x\)
\(x=\dfrac{5}{3}.\dfrac{3}{20}\)
\(x=\dfrac{1}{4}\)
Vậy \(x=\dfrac{1}{4}\)
a, \(\dfrac{11}{12}-\left(\dfrac{2}{3}+x\right)=\dfrac{2}{3}\Rightarrow x=\dfrac{11}{12}-\dfrac{2}{3}-\dfrac{2}{3}=\dfrac{-5}{12}\)
b, \(1\dfrac{1}{3}:0,8=\dfrac{2}{3}:0,1x\Rightarrow\dfrac{4}{3}:\dfrac{4}{5}=\dfrac{2}{3}:\dfrac{1}{10}x\Rightarrow\dfrac{4}{3}.\dfrac{1}{10}x=\dfrac{4}{5}.\dfrac{2}{3}\Rightarrow\dfrac{4}{3}.\dfrac{1}{10}x=\dfrac{8}{15}\Rightarrow x=\dfrac{8}{15}.\dfrac{3}{4}.10=4\)
