a: Ta có: \(\dfrac{2}{3}-1\dfrac{5}{7}x=-\dfrac{10}{3}\)
\(\Leftrightarrow x\cdot\dfrac{12}{7}=\dfrac{12}{3}\)
hay \(x=\dfrac{7}{3}\)
b: Ta có: \(\dfrac{1}{10}:\left(\dfrac{3}{4}-2x\right)^2=\dfrac{2}{5}\)
\(\Leftrightarrow\left(2x-\dfrac{3}{4}\right)^2=\dfrac{1}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{3}{4}=\dfrac{1}{2}\\2x-\dfrac{3}{4}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{5}{4}\\2x=\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{5}{8}\\2x=\dfrac{1}{8}\end{matrix}\right.\)Δ
a) \(\dfrac{2}{3}-\dfrac{12}{7}x=\dfrac{10}{3}\\ \dfrac{12}{7}x=\dfrac{2}{3}-\dfrac{10}{3}\\ \dfrac{12}{7}x=\dfrac{-8}{3}\\ x=\dfrac{-8}{3}:\dfrac{12}{7}\\ x=\dfrac{-56}{36}\)
