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\(\left(\frac{2}{5}\right)^6.\left(\frac{25}{4}\right)^2\)
\(=\left[\left(\frac{2}{5}\right)^3\right]^2.\left(\frac{25}{4}\right)^2\)
\(=\left[\left(\frac{2}{5}\right)^3.\frac{25}{4}\right]^2\)
\(=\left[\frac{8}{125}.\frac{25}{4}\right]^2\)
\(=\left(\frac{2}{5}\right)^2\)
\(=\frac{4}{25}\)
\(15\frac{1}{5}:\left(\frac{-5}{7}\right)-25\frac{1}{5}.\left(\frac{-7}{5}\right)\)
\(=15\frac{1}{5}.\frac{-7}{5}-25\frac{1}{5}.\frac{-7}{5}\)
\(=\frac{-7}{5}\left(15\frac{1}{5}-25\frac{1}{5}\right)\)
\(=\frac{-7}{5}.\left(-10\right)\)
\(=14\)
a: \(=\dfrac{5}{3}\left(-16-\dfrac{2}{7}+28+\dfrac{2}{7}\right)=\dfrac{5}{3}\cdot12=20\)
b: \(=\left(4\cdot\dfrac{3}{4}-\dfrac{1}{2}\right)\cdot\dfrac{6}{5}-17=\dfrac{1}{2}\cdot\dfrac{6}{5}-17=\dfrac{3}{5}-17=-\dfrac{82}{5}\)
c: \(=-\left(\dfrac{1}{3}\right)^{50}\cdot3^{50}-\dfrac{2}{3}\cdot\dfrac{1}{4}=-1-\dfrac{1}{6}=-\dfrac{7}{6}\)
e: \(=5.7\left(-6.5-3.5\right)=-5.7\cdot10=-57\)
Bài 1:
a) Ta có: \(25\cdot\left(\frac{-1}{5}\right)^3+\frac{1}{5}-2\cdot\left(\frac{-1}{2}\right)^2-\frac{1}{2}\)
\(=25\cdot\frac{-1}{125}+\frac{1}{5}-2\cdot\frac{1}{4}-\frac{1}{2}\)
\(=-\frac{1}{5}+\frac{1}{5}-\frac{1}{2}-\frac{1}{2}\)
\(=\frac{-2}{2}=-1\)
b) Ta có: \(35\frac{1}{6}:\left(\frac{-4}{5}\right)-46\frac{1}{6}:\left(\frac{-4}{5}\right)\)
\(=\frac{211}{6}\cdot\frac{-5}{4}-\frac{277}{6}\cdot\frac{-5}{4}\)
\(=\frac{-5}{4}\cdot\left(\frac{211}{6}-\frac{277}{6}\right)\)
\(=\frac{-5}{4}\cdot\left(-11\right)=\frac{55}{4}\)
c) Ta có: \(\left(\frac{-3}{4}+\frac{2}{5}\right):\frac{3}{7}+\left(\frac{3}{5}+\frac{-1}{4}\right):\frac{3}{7}\)
\(=\frac{-7}{20}\cdot\frac{7}{3}+\frac{7}{20}\cdot\frac{7}{3}\)
\(=\frac{7}{3}\cdot\left(-\frac{7}{20}+\frac{7}{20}\right)=\frac{7}{3}\cdot0=0\)
d) Ta có: \(\frac{7}{8}:\left(\frac{2}{9}-\frac{1}{18}\right)+\frac{7}{8}\cdot\left(\frac{1}{36}-\frac{5}{12}\right)\)
\(=\frac{7}{8}\cdot6+\frac{7}{8}\cdot\frac{-7}{18}\)
\(=\frac{7}{8}\cdot\left(6+\frac{-7}{18}\right)\)
\(=\frac{7}{8}\cdot\frac{101}{18}=\frac{707}{144}\)
e) Ta có: \(\frac{1}{6}+\frac{5}{6}\cdot\frac{3}{2}-\frac{3}{2}+1\)
\(=\frac{1}{6}+\frac{15}{12}-\frac{3}{2}+1\)
\(=\frac{2}{12}+\frac{15}{12}-\frac{18}{12}+\frac{12}{12}\)
\(=\frac{11}{12}\)
f) Ta có: \(\left(-0,75-\frac{1}{4}\right):\left(-5\right)+\frac{1}{15}-\left(-\frac{1}{5}\right):\left(-3\right)\)
\(=\left(-1\right):\left(-5\right)+\frac{1}{15}-\frac{1}{15}\)
\(=\frac{1}{5}\)
\(A=\frac{15}{34}+\frac{7}{21}+\frac{9}{34}-1\frac{15}{17}+\frac{2}{3}=\frac{15}{34}+\frac{7}{21}+\frac{9}{34}-\frac{64}{34}+\frac{14}{21}=\left(\frac{15}{34}+\frac{9}{34}-\frac{64}{34}\right)+\left(\frac{7}{21}+\frac{14}{21}\right)=\frac{30}{34}+\frac{21}{21}=\frac{15}{17}+1=\frac{32}{17}\)
Ta có: \(15\frac{1}{4}:\left(-\frac{5}{7}\right)-25\frac{1}{4}:\left(-\frac{7}{7}\right)\)
\(=\frac{61}{4}:\frac{-5}{7}+\frac{101}{4}\)
\(=\frac{61}{4}\cdot\frac{7}{-5}+\frac{101}{4}\)
\(=\frac{427}{-20}+\frac{101}{4}\)
\(=\frac{427}{-20}+\frac{-505}{-20}=\frac{-80}{-20}=4\)
15'1/4:(-5/7)-25'1/4:(-5/7)
=(15'1/4-25'1/4):(-5/7)
=-10:(-5/7)
=-10.(-7/5)
=-70/5=-14