a: Ta có: \(\dfrac{2}{3}+\dfrac{-1}{3}\cdot\dfrac{7}{8}\)
\(=\dfrac{2}{3}+\dfrac{-7}{24}\)
\(=\dfrac{16}{24}-\dfrac{7}{24}=\dfrac{9}{24}=\dfrac{3}{8}\)
b: Ta có: \(-\left(\dfrac{3}{17}+\dfrac{2}{13}\right)-\left(-\dfrac{20}{17}+\dfrac{11}{13}\right)\)
\(=\dfrac{-3}{17}-\dfrac{2}{13}+\dfrac{20}{17}-\dfrac{11}{13}\)
=0
Bài 2:
a: Ta có: \(\dfrac{2}{3}-1\dfrac{5}{7}x=-\dfrac{10}{3}\)
\(\Leftrightarrow x\cdot\dfrac{12}{7}=\dfrac{12}{3}\)
hay \(x=\dfrac{12}{3}:\dfrac{12}{7}=\dfrac{7}{3}\)
b: Ta có: \(\dfrac{1}{10}:\left(\dfrac{3}{4}-2x\right)^2=0.4\)
\(\Leftrightarrow\left(\dfrac{3}{4}-2x\right)^2=\dfrac{1}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{3}{4}=\dfrac{1}{4}\\2x-\dfrac{3}{4}=-\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=1\\2x=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{1}{4}\end{matrix}\right.\)
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: \(\dfrac{1}{10}:\left(\dfrac{3}{4}-2x\right)^2=\dfrac{2}{5}\)
\(\Leftrightarrow\left(2x-\dfrac{3}{4}\right)^2=\dfrac{1}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{3}{4}=\dfrac{1}{2}\\2x-\dfrac{3}{4}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{5}{4}\\2x=\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{5}{8}\\2x=\dfrac{1}{8}\end{matrix}\right.\)
