Question

Answer 1

a) Ta có: \(\left(\dfrac{2}{5}-x\right):1\dfrac{1}{3}+\dfrac{1}{2}=-4\)

\(\Leftrightarrow\left(\dfrac{2}{5}-x\right):\dfrac{4}{3}=-4-\dfrac{1}{2}=-\dfrac{9}{2}\)

\(\Leftrightarrow\left(\dfrac{2}{5}-x\right)=\dfrac{-9}{2}\cdot\dfrac{4}{3}=-\dfrac{36}{6}=-6\)

hay \(x=\dfrac{32}{5}\)

Vậy: \(x=\dfrac{32}{5}\)

b) Ta có: \(\left(-3+\dfrac{3}{x}-\dfrac{1}{3}\right):\left(1+\dfrac{2}{5}+\dfrac{2}{3}\right)=-\dfrac{5}{4}\)

\(\Leftrightarrow\left(\dfrac{3}{x}-\dfrac{10}{3}\right):\dfrac{31}{15}=\dfrac{-5}{4}\)

\(\Leftrightarrow\dfrac{3}{x}-\dfrac{10}{3}=\dfrac{-31}{12}\)

\(\Leftrightarrow\dfrac{3}{x}=\dfrac{3}{4}\)

hay x=4

Vậy: x=4