a, \(\Leftrightarrow\dfrac{12x}{7}=\dfrac{2}{3}+\dfrac{10}{3}=4\Rightarrow12x=28\Leftrightarrow x=\dfrac{28}{12}=\dfrac{7}{3}\)
b, \(\Leftrightarrow\left(2x-\dfrac{3}{4}\right)^2=\dfrac{1}{10}:\dfrac{2}{5}=\dfrac{1}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{3}{4}=\dfrac{1}{2}\\2x-\dfrac{3}{4}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{8}\\x=\dfrac{1}{8}\end{matrix}\right.\)
\(a,\dfrac{2}{3}-1\dfrac{5}{7}x=-\dfrac{10}{3}\)
\(\dfrac{12}{7}x=\dfrac{2}{3}-\dfrac{-10}{3}\)
\(\dfrac{12}{7}x=4\)
\(x=4:\dfrac{12}{7}\)
\(x=4.\dfrac{7}{12}\)
\(x=\dfrac{7}{3}\)
\(b,\dfrac{1}{10}:\left(\dfrac{3}{4}-2x\right)^2=0,4\)
\(\dfrac{1}{10}:\left(\dfrac{3}{4}-2x\right)^2=\dfrac{2}{5}\)
\(\left(\dfrac{3}{4}-2x\right)^2=\dfrac{1}{10}:\dfrac{2}{5}\)
\(\left(\dfrac{3}{4}-2x\right)^2=\dfrac{1}{10}.\dfrac{5}{2}\)
\(\left(\dfrac{3}{4}-2x\right)^2=\dfrac{1}{4}\)
\(\left(\dfrac{3}{4}-2x\right)^2=\pm\left(\dfrac{1}{2}\right)^2\)
\(\Rightarrow\dfrac{3}{4}-2x=\pm\dfrac{1}{2}\)
\(TH1:\dfrac{3}{4}-2x=\dfrac{1}{2}\)
\(2x=\dfrac{3}{4}-\dfrac{1}{2}\)
\(2x=\dfrac{1}{4}\)
\(x=\dfrac{1}{4}:2\)
\(x=\dfrac{1}{4}.\dfrac{1}{2}\)
\(x=\dfrac{1}{8}\)
\(TH2:\dfrac{3}{4}-2x=\dfrac{-1}{2}\)
\(2x=\dfrac{3}{4}-\dfrac{-1}{2}\)
\(2x=\dfrac{5}{4}\)
\(x=\dfrac{5}{4}:2\)
\(x=\dfrac{5}{4}.\dfrac{1}{2}\)
\(x=\dfrac{5}{8}\)
\(Vậy\) \(x\in\left\{\dfrac{1}{8};\dfrac{5}{8}\right\}\)
