a) 3.\(\dfrac{5}{4}\) - \(\dfrac{\left(-3\right)^2}{4}\)
= \(\dfrac{15}{4}-\dfrac{9}{4}\)
= \(\dfrac{6}{4}=\dfrac{3}{2}\)
b) \(\dfrac{6}{7}.\dfrac{8}{13}+\dfrac{6}{13}.\dfrac{9}{7}-\dfrac{3}{13}.\dfrac{6}{7}\)
= \(\dfrac{6}{7}.\dfrac{8}{13}+\dfrac{6}{7}.\dfrac{9}{13}-\dfrac{3}{13}.\dfrac{6}{7}\)
= \(\dfrac{6}{7}.\left(\dfrac{8}{13}+\dfrac{9}{13}-\dfrac{3}{13}\right)\)
= \(\dfrac{6}{7}.\dfrac{14}{13}=\dfrac{12}{13}\)
c) \(3\dfrac{1}{7}:9\dfrac{3}{4}+3\dfrac{6}{7}:9\dfrac{3}{4}\)
= \(\left(3\dfrac{1}{7}+3\dfrac{6}{7}\right):9\dfrac{3}{4}\)
= 7 : \(\dfrac{39}{4}\) = \(7.\dfrac{4}{39}\) = \(\dfrac{28}{39}\)
d) \(\left(0,75-\dfrac{1}{4}\right):\dfrac{3}{8}\)
= \(\left(\dfrac{3}{4}-\dfrac{1}{4}\right):\dfrac{3}{8}\)
= \(\dfrac{1}{2}:\dfrac{3}{8}\) = \(\dfrac{1}{2}.\dfrac{8}{3}=\dfrac{4}{3}\)
Bài 6:
a) Ta có: \(\left(\dfrac{2}{3}-\dfrac{1}{2}x\right):\dfrac{2}{3}=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{2}{3}-\dfrac{1}{2}x=\dfrac{3}{4}\cdot\dfrac{2}{3}=\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{1}{2}x=\dfrac{2}{3}-\dfrac{1}{2}=\dfrac{1}{6}\)
hay \(x=\dfrac{1}{3}\)
b) Ta có: \(\dfrac{4}{5}+\dfrac{3}{5}:x=1\dfrac{1}{5}\)
\(\Leftrightarrow\dfrac{3}{5}:x=\dfrac{4}{5}-\dfrac{6}{5}=-\dfrac{2}{5}\)
hay \(x=\dfrac{3}{5}:\dfrac{-2}{5}=\dfrac{3}{5}\cdot\dfrac{5}{-2}=\dfrac{-3}{2}\)
c) Ta có: \(\dfrac{3}{x+5}=15\%\)
\(\Leftrightarrow x+5=20\)
hay x=15
d) Ta có: \(6x-7.2x=14.4\)
\(\Leftrightarrow-1.2x=14.4\)
hay x=-12
a) Ta có: \(3\cdot\dfrac{5}{4}-\dfrac{\left(-3\right)^2}{4}\)
\(=\dfrac{15}{4}-\dfrac{9}{4}=\dfrac{6}{4}=\dfrac{3}{2}\)
b) Ta có: \(\dfrac{6}{7}\cdot\dfrac{8}{13}+\dfrac{6}{13}\cdot\dfrac{9}{7}-\dfrac{3}{13}\cdot\dfrac{6}{7}\)
\(=\dfrac{6}{7}\cdot\dfrac{8}{13}+\dfrac{6}{7}\cdot\dfrac{9}{13}-\dfrac{6}{7}\cdot\dfrac{3}{13}\)
\(=\dfrac{6}{7}\left(\dfrac{8}{13}+\dfrac{9}{13}-\dfrac{3}{13}\right)\)
\(=\dfrac{6}{7}\cdot\dfrac{14}{13}=\dfrac{12}{13}\)
c) Ta có: \(3\dfrac{1}{7}:9\dfrac{3}{4}+3\dfrac{6}{7}:9\dfrac{3}{4}\)
\(=\dfrac{22}{7}:\dfrac{39}{4}+\dfrac{27}{7}:\dfrac{39}{4}\)
\(=\dfrac{22}{7}\cdot\dfrac{4}{39}+\dfrac{27}{7}\cdot\dfrac{4}{39}\)
\(=\dfrac{4}{39}\cdot7=\dfrac{28}{39}\)