\(a,\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{1}{10}\Leftrightarrow\dfrac{2}{3}x=\dfrac{1}{10}+\dfrac{1}{2}\Leftrightarrow\dfrac{2}{3}x=\dfrac{2}{3}\Leftrightarrow x=\dfrac{2}{3}:\dfrac{2}{3}=1\)\(b,5\dfrac{4}{7}:x=13\Leftrightarrow\dfrac{39}{7}:x=13\Leftrightarrow x=\dfrac{39}{7}:13=\dfrac{3}{7}\)\(c,\left(2\dfrac{4}{5}x-50\right):\dfrac{2}{3}=51\Leftrightarrow\left(\dfrac{14}{5}x-50\right).\dfrac{3}{2}=51\Leftrightarrow\dfrac{21}{5}x-75=51\Leftrightarrow\dfrac{21}{5}x=51+75=126\Leftrightarrow x=126:\dfrac{21}{5}=30\)
d,\(\left(x+\dfrac{1}{2}\right).\left(\dfrac{2}{3}-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\\dfrac{2}{3}-2x=0\end{matrix}\right.\left[{}\begin{matrix}x=\dfrac{-1}{2}\\2x=\dfrac{2}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
a, \(\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{1}{10}\)
=> \(\dfrac{2}{3}x=\dfrac{1}{10}+\dfrac{1}{2}=\dfrac{3}{5}\)
=> \(x=\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{9}{10}\)
b, Dễ ==> tự lm
c,( \(2\dfrac{4}{5}x-50\)) \(:\dfrac{2}{3}=51\)
=> \(2\dfrac{4}{5}x-50=51\cdot\dfrac{2}{3}=34\)
=> \(\dfrac{14}{5}x=34+50=84\)
=> \(x=84:\dfrac{14}{5}=30\)
d, \(\left(x+\dfrac{1}{2}\right)\left(\dfrac{2}{3}-2x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\\dfrac{2}{3}-2x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\-2x=-\dfrac{2}{3}\Rightarrow x=\dfrac{1}{3}\end{matrix}\right.\)
e, \(\dfrac{2}{3}-\dfrac{1}{2}x=\dfrac{5}{12}\)
=> \(-\dfrac{1}{2}x=\dfrac{5}{12}-\dfrac{2}{3}=-\dfrac{1}{4}\)
=> \(x=-\dfrac{1}{4}:\left(-\dfrac{1}{2}\right)=\dfrac{1}{2}\)
e,\(\dfrac{2}{3}-\dfrac{1}{2}x=\dfrac{5}{12}\Leftrightarrow\dfrac{1}{2}x=\dfrac{2}{3}-\dfrac{5}{12}=\dfrac{1}{4}\Leftrightarrow x=\dfrac{1}{4}:\dfrac{1}{2}=\dfrac{1}{2}\)
a) \(\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{1}{10}\)
\(\Rightarrow\dfrac{2}{3}x=\dfrac{1}{10}+\dfrac{1}{2}=\dfrac{3}{5}\)
\(\Rightarrow x=\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{3}{5}\cdot\dfrac{3}{2}=\dfrac{9}{10}\)
Vậy \(x=\dfrac{9}{10}\).
b) \(5\dfrac{4}{7}:x=13\)
\(\Leftrightarrow\dfrac{39}{7}:x=13\)
\(\Rightarrow x=\dfrac{39}{7}:13=\dfrac{39}{7}\cdot\dfrac{1}{13}=\dfrac{3}{7}\)
Vậy \(x=\dfrac{3}{7}\).
c) \(\left(2\dfrac{4}{5}x-50\right):\dfrac{2}{3}=51\)
\(\Leftrightarrow\left(\dfrac{14}{5}x-50\right):\dfrac{2}{3}=51\)
\(\Rightarrow\dfrac{14}{5}x-50=51\cdot\dfrac{2}{3}=34\)
\(\Rightarrow\dfrac{14}{5}x=34+50=84\)
\(\Rightarrow x=84:\dfrac{14}{5}=84\cdot\dfrac{5}{14}=30\)
Vậy \(x=30\).
d) \(\left(x+\dfrac{1}{2}\right)\left(\dfrac{2}{3}-2x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\\dfrac{2}{3}-2x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{-1}{2};\dfrac{1}{3}\right\}\).
e) \(\dfrac{2}{3}-\dfrac{1}{2}x=\dfrac{5}{12}\)
\(\Rightarrow\dfrac{1}{2}x=\dfrac{2}{3}-\dfrac{5}{12}=\dfrac{1}{4}\)
\(\Rightarrow x=\dfrac{1}{4}:\dfrac{1}{2}=\dfrac{1}{4}\cdot2=\dfrac{1}{2}\)
Vậy \(x=\dfrac{1}{2}\).
