Bài 1:
a) \(3\dfrac{4}{5}.\left(-1\dfrac{1}{2}\right)+3\dfrac{4}{5}.2\dfrac{1}{2}-1\dfrac{4}{7}\)
\(=\dfrac{19}{5}.\left(\dfrac{-3}{2}+\dfrac{5}{2}\right)-\dfrac{11}{7}\)
\(=\dfrac{19}{5}.1-\dfrac{11}{7}\)
\(=\dfrac{19}{5}-\dfrac{11}{7}\)
\(=\dfrac{78}{35}\)
b) \(\dfrac{3}{4}:\left(-2\dfrac{1}{2}\right)-\dfrac{4}{5}:\left(-2\dfrac{1}{2}\right)+1\dfrac{1}{4}:\left(-2\dfrac{1}{2}\right)\)
\(=\left(\dfrac{3}{4}-\dfrac{4}{5}+\dfrac{5}{4}\right):\dfrac{-5}{2}\)
\(=\dfrac{6}{5}:\dfrac{-5}{2}\)
\(=\dfrac{-12}{25}\)
c) \(\dfrac{4}{5}-\left[\dfrac{3}{5}.\left(\dfrac{-2}{3}-1\dfrac{1}{3}\right)+\dfrac{4}{7}\right]\)
\(=\dfrac{4}{5}-\left[\dfrac{3}{5}.\left(\dfrac{-2}{3}-\dfrac{4}{3}\right)+\dfrac{4}{7}\right]\)
\(=\dfrac{4}{5}-\left[\dfrac{3}{5}.-2+\dfrac{4}{7}\right]\)
\(=\dfrac{4}{5}-\left[\dfrac{-6}{5}+\dfrac{4}{7}\right]\)
\(=\dfrac{4}{5}-\left(\dfrac{-22}{35}\right)\)
\(=\dfrac{10}{7}\)
Bài 2:
a) \(\left(\dfrac{3}{4}-x\right).\dfrac{1}{2}-\dfrac{4}{5}=1\dfrac{1}{5}\)
\(\left(\dfrac{3}{4}-x\right).\dfrac{1}{2}=\dfrac{6}{5}+\dfrac{4}{5}\)
\(\left(\dfrac{3}{4}-x\right).\dfrac{1}{2}=2\)
\(\dfrac{3}{4}-x=2:\dfrac{1}{2}\)
\(\dfrac{3}{4}-x=4\)
\(x=\dfrac{3}{4}-4\)
\(x=\dfrac{-13}{4}\)
b) \(\dfrac{3}{4}-\dfrac{1}{4}.x=2\dfrac{1}{3}\)
\(\dfrac{1}{4}.x=\dfrac{3}{4}-\dfrac{7}{3}\)
\(\dfrac{1}{4}.x=\dfrac{-19}{12}\)
\(x=\dfrac{-19}{12}:\dfrac{1}{4}\)
\(x=\dfrac{-19}{3}\)
Bài 2:
c) \(\left(\dfrac{4}{8}+x\right):\dfrac{2}{5}-\dfrac{1}{3}=1\dfrac{2}{3}\)
\(\left(\dfrac{1}{2}+x\right):\dfrac{2}{5}=\dfrac{5}{3}+\dfrac{1}{3}\)
\(\left(\dfrac{1}{2}+x\right):\dfrac{2}{5}=2\)
\(\dfrac{1}{2}+x=2.\dfrac{2}{5}\)
\(\dfrac{1}{2}+x=\dfrac{4}{5}\)
\(x=\dfrac{4}{5}-\dfrac{1}{2}\)
\(x=\dfrac{3}{10}\)
Bài 3:
Ta có: \(A=\dfrac{4}{5}.a+1\dfrac{1}{5}.a-\dfrac{3}{4}.a\)
Ta thay: \(a=\dfrac{-2}{3}\)
Ta có:
\(A=\dfrac{4}{5}.\dfrac{-2}{3}+1\dfrac{1}{5}.\dfrac{-2}{3}-\dfrac{3}{4}.\dfrac{-2}{3}\)
\(A=\dfrac{-2}{3}.\left(\dfrac{4}{5}+\dfrac{6}{5}-\dfrac{3}{4}\right)\)
\(A=\dfrac{-2}{3}.\dfrac{5}{4}\)
\(A=\dfrac{-5}{6}\)