Bài 1:
a) \(\dfrac{3}{8}+\dfrac{4}{5}=\dfrac{15}{40}+\dfrac{32}{40}=\dfrac{47}{40}\)
b) \(7\dfrac{2}{9}-\left(4\dfrac{3}{4}+2\dfrac{2}{9}\right)=\dfrac{65}{9}-\dfrac{19}{4}-\dfrac{20}{4}=5-\dfrac{19}{4}=\dfrac{1}{4}\)
c) \(\dfrac{6}{11}\cdot\dfrac{-15}{13}+\dfrac{6}{11}\cdot\dfrac{-2}{13}-\dfrac{6}{11}\cdot\dfrac{9}{13}=\dfrac{6}{11}\left(\dfrac{-15}{13}-\dfrac{2}{13}-\dfrac{9}{13}\right)=\dfrac{6}{11}\cdot\left(-2\right)=\dfrac{-12}{11}\)
Bài 2:
a) Ta có: \(x:\dfrac{2}{5}=\dfrac{-10}{7}\)
nên \(x=\dfrac{-10}{7}\cdot\dfrac{2}{5}=\dfrac{-20}{35}=\dfrac{-4}{7}\)
b) Ta có: \(\dfrac{-3}{4}x+\dfrac{5}{12}=\dfrac{1}{6}\)
nên \(\dfrac{-3}{4}x=\dfrac{1}{6}-\dfrac{5}{12}=\dfrac{-1}{4}\)
hay \(x=\dfrac{1}{4}:\dfrac{3}{4}=\dfrac{1}{3}\)
c) Ta có: \(\dfrac{8}{3}+\dfrac{2}{3}\left(4x-1\right)=7\dfrac{1}{3}\)
\(\Leftrightarrow\dfrac{8}{3}+\dfrac{8}{3}x-\dfrac{2}{3}-\dfrac{22}{3}=0\)
\(\Leftrightarrow\dfrac{8}{3}x=\dfrac{16}{3}\)
hay x=2
Bài 1:
d) Ta có: \(40\%-4\dfrac{2}{5}\cdot2021^0+\left|-1.5\right|:\dfrac{3}{8}\)
\(=\dfrac{2}{5}-\dfrac{22}{5}+\dfrac{3}{2}\cdot\dfrac{8}{3}\)
\(=-4+4=0\)
Bài 2:
d) Ta có: \(\left(\dfrac{3}{5}-x\right)^2-1.6=\dfrac{9}{25}\)
\(\Leftrightarrow\left(x-\dfrac{3}{5}\right)^2=\dfrac{9}{25}+\dfrac{8}{5}=\dfrac{49}{25}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{5}=\dfrac{7}{5}\\x-\dfrac{3}{5}=-\dfrac{7}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{-4}{5}\end{matrix}\right.\)
