Bài 1:
a: \(A=\left(-3,2\right)\cdot\dfrac{-15}{64}+\left(0,8-2\dfrac{4}{5}\right):3\dfrac{2}{3}\)
\(=\dfrac{16}{5}\cdot\dfrac{15}{64}+\left(0,8-2-0,8\right):\dfrac{11}{3}\)
\(=\dfrac{3}{4}-2\cdot\dfrac{3}{11}=\dfrac{3}{4}-\dfrac{6}{11}=\dfrac{33-24}{44}=\dfrac{9}{44}\)
b: \(B=\dfrac{4}{7}\cdot\dfrac{8}{-13}+\dfrac{4}{7}\cdot\dfrac{-2}{13}+\left(1\dfrac{3}{13}\right)\cdot\dfrac{-4}{7}\)
\(=\dfrac{4}{7}\left(-\dfrac{8}{13}-\dfrac{2}{13}-1-\dfrac{3}{13}\right)\)
\(=\dfrac{4}{7}\cdot\left(-2\right)=-\dfrac{8}{7}\)
Bài 2:
a: \(\left(2\dfrac{4}{5}x-0,2\right):\dfrac{4}{5}=\dfrac{7}{8}\)
=>\(\left(2,8x-0,2\right)=\dfrac{7}{8}\cdot\dfrac{4}{5}=\dfrac{7}{2}\cdot\dfrac{1}{5}=\dfrac{7}{10}=0,7\)
=>2,8x=0,7+0,2=0,9
=>28x=9
=>\(x=\dfrac{9}{28}\)
b: \(\dfrac{1}{4}+\dfrac{1}{3}:\left|2x-1\right|=\dfrac{11}{12}\)
=>\(\dfrac{1}{3}:\left|2x-1\right|=\dfrac{11}{12}-\dfrac{1}{4}=\dfrac{11}{12}-\dfrac{3}{12}=\dfrac{8}{12}=\dfrac{2}{3}\)
=>\(\left|2x-1\right|=\dfrac{1}{3}:\dfrac{2}{3}=\dfrac{1}{2}\)
=>\(\left[{}\begin{matrix}2x-1=\dfrac{1}{2}\\2x-1=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{3}{2}\\2x=\dfrac{1}{2}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{1}{4}\end{matrix}\right.\)
c: \(\dfrac{3}{5}-\left(2\dfrac{1}{5}-x\right)^2=\dfrac{6}{25}\)
=>\(\dfrac{3}{5}-\left(x-\dfrac{7}{5}\right)^2=\dfrac{6}{25}\)
=>\(\left(x-\dfrac{7}{5}\right)^2=\dfrac{3}{5}-\dfrac{6}{25}=\dfrac{15}{25}-\dfrac{6}{25}=\dfrac{9}{25}\)
=>\(\left[{}\begin{matrix}x-\dfrac{7}{5}=\dfrac{3}{5}\\x-\dfrac{7}{5}=-\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{4}{5}\end{matrix}\right.\)
