Câu 1:
a) \(\dfrac{-7}{5}+\dfrac{4}{5}=\dfrac{-7+4}{5}=\dfrac{-3}{5}\)
b) Ta có: \(\dfrac{-3}{5}\cdot\dfrac{3}{7}+\dfrac{-3}{5}\cdot\dfrac{4}{7}+2\dfrac{1}{3}\)
\(=\dfrac{-3}{5}\cdot\left(\dfrac{3}{7}+\dfrac{4}{7}\right)+\dfrac{7}{3}\)
\(=\dfrac{-3}{5}+\dfrac{7}{3}\)
\(=\dfrac{-9+35}{15}=\dfrac{26}{15}\)
Câu 1:
c) Ta có: \(\dfrac{2}{5}-\dfrac{3}{5}:3-\dfrac{7}{16}\cdot\left(-2\right)^2\)
\(=\dfrac{2}{5}-\dfrac{3}{5}\cdot\dfrac{1}{3}-\dfrac{7}{16}\cdot4\)
\(=\dfrac{2}{5}-\dfrac{1}{5}-\dfrac{7}{4}\)
\(=\dfrac{1}{5}-\dfrac{7}{4}=\dfrac{4-35}{20}=\dfrac{-31}{20}\)
Câu 2:
a) Ta có: \(\dfrac{2}{3}-x=\dfrac{5}{4}\)
\(\Leftrightarrow x=\dfrac{2}{3}-\dfrac{5}{4}=\dfrac{8}{12}-\dfrac{15}{12}=\dfrac{-7}{12}\)
Vậy: \(x=-\dfrac{7}{12}\)
Câu 2:
b) Ta có: \(\dfrac{4}{5}+\dfrac{5}{7}:x=\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{5}{7}:x=\dfrac{1}{6}-\dfrac{4}{5}=\dfrac{5-24}{30}=\dfrac{-19}{30}\)
hay \(x=\dfrac{-150}{133}\)
Vậy: \(x=\dfrac{-150}{133}\)
