Câu 2:
a: \(x+\dfrac{3}{4}=\dfrac{1}{2}\)
=>\(x=\dfrac{1}{2}-\dfrac{3}{4}\)
=>\(x=\dfrac{2}{4}-\dfrac{3}{4}=-\dfrac{1}{4}\)
b: \(\dfrac{1}{2}x-\dfrac{3}{4}=\dfrac{5}{4}\)
=>\(\dfrac{1}{2}x=\dfrac{5}{4}+\dfrac{3}{4}=\dfrac{8}{4}=2\)
=>\(x=2\cdot2=4\)
c: \(\dfrac{x}{56}=-\dfrac{7}{8}\)
=>\(x=-\dfrac{7}{8}\cdot56=-49\)
Câu 1:
a: \(\dfrac{3}{5}+\dfrac{7}{12}\cdot\dfrac{6}{7}\)
\(=\dfrac{3}{5}+\dfrac{6}{12}\)
\(=\dfrac{3}{5}+\dfrac{1}{2}=\dfrac{6+5}{10}=\dfrac{11}{10}\)
b: \(\dfrac{1}{3}\cdot\dfrac{4}{5}+\dfrac{1}{3}\cdot\dfrac{6}{5}+\dfrac{2}{3}\)
\(=\dfrac{1}{3}\left(\dfrac{4}{5}+\dfrac{6}{5}\right)+\dfrac{2}{3}=\dfrac{1}{3}\cdot2+\dfrac{2}{3}=\dfrac{4}{3}\)
c: \(\left(-2\right)^3\cdot\dfrac{-1}{24}+\left(1\dfrac{2}{3}+2\dfrac{3}{5}\right):\dfrac{4}{15}\)
\(=\left(-8\right)\cdot\dfrac{-1}{24}+\left(3+\dfrac{2}{3}+\dfrac{3}{5}\right)\cdot\dfrac{15}{4}\)
\(=\dfrac{1}{3}+\dfrac{45+10+9}{15}\cdot\dfrac{15}{4}\)
\(=\dfrac{1}{3}+\dfrac{64}{4}=\dfrac{1}{3}+16=\dfrac{49}{3}\)
