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16 tháng 7 2016

\(P=\left(-0,5-\frac{3}{5}\right):\left(-3\right)+\frac{1}{3}-\left(-\frac{1}{6}\right):\left(-2\right)\)

\(P=\left(-1,1\right):\left(-3\right)+\frac{1}{3}+\frac{1}{6}:\left(-2\right)\)

\(P=\frac{11}{30}+\frac{1}{3}+\left(-\frac{1}{12}\right)\)

\(P=\frac{37}{60}\)

\(Q=\left(\frac{2}{25}-1,008\right):\frac{4}{7}:\left[\left(3\frac{1}{4}-6\frac{5}{9}\right).2\frac{2}{17}\right]\)

\(Q=\left(-0,928\right):\frac{4}{7}:\left[\left(-\frac{119}{36}\right).2\frac{2}{17}\right]\)

\(Q=\left(-1,624\right):\left(-\frac{245}{36}\right)\)

\(Q=\frac{1044}{4375}\)

\(P=\left(\dfrac{-1}{2}-\dfrac{3}{5}\right):\left(-3\right)+\dfrac{1}{3}-\dfrac{1}{6}:2\)

\(=\left(\dfrac{1}{2}+\dfrac{3}{5}\right):3+\dfrac{1}{3}-\dfrac{1}{12}\)

\(=\dfrac{11}{10}\cdot\dfrac{1}{3}+\dfrac{1}{4}\)

\(=\dfrac{11}{30}+\dfrac{1}{4}=\dfrac{22}{60}+\dfrac{15}{60}=\dfrac{37}{60}\)

\(Q=\left(\dfrac{2}{25}-\dfrac{126}{125}\right)\cdot\dfrac{7}{4}:\left[\dfrac{-119}{36}\cdot\dfrac{36}{17}\right]\)

\(=\dfrac{-116}{125}\cdot\dfrac{7}{4}:\left(-7\right)\)

\(=\dfrac{116}{125}\cdot\dfrac{7}{4}\cdot\dfrac{1}{7}=\dfrac{29}{125}\)

19 tháng 9 2015

ảo                         

10 tháng 6 2018

\(A=2\frac{1}{2}\)

23 tháng 9 2020

a) \(\left(-\frac{3}{4}\right)^2:\left(\frac{5}{4}\right)^2+14,7-1\frac{9}{25}\)

\(=\left[\left(-\frac{3}{4}\right):\frac{5}{4}\right]^2+\frac{147}{10}-\frac{34}{25}\)

\(=\left[\left(-\frac{3}{4}\right)\cdot\frac{4}{5}\right]^2+\frac{147}{10}-\frac{34}{25}\)

\(=\left(-\frac{3}{5}\right)^2+\frac{147}{10}-\frac{34}{25}=\frac{9}{25}+\frac{147}{10}-\frac{34}{25}=\left(\frac{9}{25}-\frac{34}{25}\right)+\frac{147}{10}=-1+\frac{147}{10}=\frac{137}{10}\)

b) \(\left(2\frac{1}{3}-1,5\right):\left(-6\frac{1}{6}+5\frac{1}{2}\right)+2,75\)

\(=\left(\frac{7}{3}-\frac{3}{2}\right):\left(-\frac{37}{6}+\frac{11}{2}\right)+\frac{11}{4}\)

\(=\frac{5}{6}:\left(-\frac{2}{3}\right)+\frac{11}{4}=\frac{5}{6}\cdot\left(-\frac{3}{2}\right)+\frac{11}{4}=-\frac{5}{4}+\frac{11}{4}=\frac{3}{2}\)

23 tháng 9 2020

                                                                Bài giải

\(a,\text{ }\left(-\frac{3}{4}\right)^2\text{ : }\left(\frac{5}{4}\right)^2+14,7-1\frac{9}{25}\)

\(=\frac{9}{16}\text{ : }\frac{25}{16}+\frac{147}{10}-\frac{34}{25}\)

\(=\frac{18}{50}+\frac{735}{50}-\frac{68}{50}\)

\(=\frac{685}{50}=\frac{137}{10}\)