câu 3 ,4 ,5
Bài 3:
a) Ta có: \(\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{1}{10}\)
\(\Leftrightarrow\dfrac{2}{3}x=\dfrac{3}{5}\)
hay \(x=\dfrac{9}{10}\)
Vậy: \(x=\dfrac{9}{10}\)
b) Ta có: \(5\dfrac{4}{7}:x=13\)
\(\Leftrightarrow\dfrac{39}{7}:x=13\)
hay \(x=\dfrac{3}{7}\)
Vậy: \(x=\dfrac{3}{7}\)
c) Ta có: \(\left(x+\dfrac{1}{2}\right)\left(\dfrac{2}{3}-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\\dfrac{2}{3}-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\2x=\dfrac{2}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{-\dfrac{1}{2};\dfrac{1}{3}\right\}\)
d) Ta có: \(\dfrac{2}{3}x-\dfrac{1}{2}x=\dfrac{5}{12}\)
\(\Leftrightarrow\dfrac{1}{6}x=\dfrac{5}{12}\)
hay \(x=\dfrac{5}{2}\)
Vậy: \(x=\dfrac{5}{2}\)
e) Ta có: \(\left(x\cdot6\dfrac{2}{7}+\dfrac{3}{7}\right)\cdot2\dfrac{1}{5}-\dfrac{3}{7}=-2\)
\(\Leftrightarrow\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\cdot\dfrac{11}{5}=-\dfrac{11}{7}\)
\(\Leftrightarrow\dfrac{44}{7}x+\dfrac{3}{7}=\dfrac{-5}{7}\)
\(\Leftrightarrow\dfrac{44}{7}x=-\dfrac{8}{7}\)
hay \(x=-\dfrac{2}{11}\)
Vậy: \(x=-\dfrac{2}{11}\)
