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9 tháng 8 2016

a)\(25\frac{3}{5}:\left(\frac{-2}{3}\right)-15\frac{3}{5}:\left(\frac{-2}{3}\right)\)

\(=\left(25\frac{3}{5}-15\frac{3}{5}\right):\left(-\frac{2}{3}\right)\)

\(=10:\left(\frac{-2}{3}\right)\)

\(=-15\)

b)\(9.\left(\frac{-2}{3}\right)^3+\frac{1}{2}:5\)

\(=9.\frac{-8}{27}+\frac{1}{10}\)

\(=\frac{-8}{3}+\frac{1}{10}\)

\(=\frac{-77}{30}\)

c)\(\left[10\left(\frac{-1}{5}\right)^2+5\left(\frac{-1}{5}\right)+1\right]:\left(\frac{-1}{5}-1\right)\)

\(=\frac{2}{5}:\left(\frac{-6}{5}\right)\)

\(=\frac{-1}{3}\)

 

9 tháng 8 2016

\(a.25\frac{3}{5}:\left(-\frac{2}{3}\right)-15\frac{3}{5}:\left(-\frac{2}{3}\right)\)

\(=\frac{128}{5}:\left(-\frac{2}{3}\right)-\frac{75}{5}:\left(-\frac{2}{3}\right)\)

\(=\left(-\frac{192}{5}\right)-\left(-\frac{117}{5}\right)\)

\(=\frac{\left(-192\right)-\left(-117\right)}{5}\)

\(=-15\)

\(b.9.\left(-\frac{2}{3}\right)^3+\frac{1}{2}:5\)

\(=9.\left(-\frac{8}{27}\right)+\frac{1}{2}:5\)

\(=-\frac{8}{3}+\frac{1}{10}\)

\(=-\frac{77}{30}\)

\(c.\left[10\left(\frac{-1}{5}\right)^2+5\left(\frac{-1}{5}\right)+1\right]:\left(\frac{-1}{5}-1\right)\)

\(=\left[10\left(\frac{-1}{25}\right)+5\left(\frac{-1}{5}\right)+1\right]:\left(\frac{-1}{5}-1\right)\)

\(=\left[\frac{-2}{5}+\left(-1\right)+1\right]:\left(-\frac{6}{5}\right)\)

\(=\left(-\frac{2}{5}\right):\left(-\frac{6}{5}\right)\)

\(=\frac{1}{3}\)

 

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: \(=5.7\left(-6.5-3.5\right)=-5.7\cdot10=-57\)

HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023

a)\({\left( {\frac{2}{5} + \frac{1}{2}} \right)^2} = {\left( {\frac{4}{{10}} + \frac{5}{{10}}} \right)^2} = {\left( {\frac{9}{{10}}} \right)^2} = \frac{{81}}{{100}}\);               

 b)\({\left( {0,75 - 1\frac{1}{2}} \right)^3} = {\left( {\frac{3}{4} - \frac{3}{2}} \right)^3} = {\left( {\frac{3}{4} - \frac{6}{4}} \right)^3} = {\left( { - \frac{3}{4}} \right)^3} = \frac{{ - 27}}{{64}};\)

c)

\(\begin{array}{l}{\left( {\frac{3}{5}} \right)^{15}}:{\left( {0,36} \right)^5} = {\left( {\frac{3}{5}} \right)^{15}}:{\left( {\frac{9}{{25}}} \right)^5}\\ = {\left( {\frac{3}{5}} \right)^{15}}:{\left[ {{{\left( {\frac{3}{5}} \right)}^2}} \right]^5} = {\left( {\frac{3}{5}} \right)^{15}}:{\left( {\frac{3}{5}} \right)^{10}} = {\left( {\frac{3}{5}} \right)^5}\end{array}\)

d) \({\left( {1 - \frac{1}{3}} \right)^8}:{\left( {\frac{4}{9}} \right)^3} = {\left( {\frac{3}{3} - \frac{1}{3}} \right)^8}:{\left( ({\frac{2}{3}})^2 \right)^3}\\= {\left( {\frac{2}{3}} \right)^8}:{\left( {\frac{2}{3}} \right)^6} = {\left( {\frac{2}{3}} \right)^{8-6}}\\= {\left( {\frac{2}{3}} \right)^2} = \frac{4}{9}\)

HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023

a)

\(\begin{array}{l}\left( {\frac{3}{4}:1\frac{1}{2}} \right) - \left( {\frac{5}{6}:\frac{1}{3}} \right)\\ = \left( {\frac{3}{4}:\frac{3}{2}} \right) - \left( {\frac{5}{6}.3} \right)\\ = \left( {\frac{3}{4}.\frac{2}{3}} \right) - \frac{5}{2}\\ = \frac{1}{2} - \frac{5}{2}\\ = \frac{-4}{2}\\= - 2.\end{array}\)                         

b)

\(\begin{array}{l}\left[ {\left( {\frac{{ - 1}}{5}} \right):\frac{1}{{10}}} \right] - \frac{5}{7}.\left( {\frac{2}{3} - \frac{1}{5}} \right)\\ = \left( {\frac{{ - 1}}{5}} \right).10 - \frac{5}{7}.\left( {\frac{{10}}{{15}} - \frac{3}{{15}}} \right)\\ =  - 2 - \frac{5}{7}.\frac{7}{{15}}\\ =  - 2 - \frac{1}{3}\\ = \frac{{ - 6}}{3} - \frac{1}{3}\\ = \frac{{ - 7}}{3}\end{array}\)

c)

\(\begin{array}{l}\left( { - 0,4} \right) + 2\frac{2}{5}.{\left[ {\left( {\frac{{ - 2}}{3}} \right) + \frac{1}{2}} \right]^2}\\ = \left( { - \frac{2}{5}} \right) + \frac{{12}}{5}.{\left[ {\left( {\frac{{ - 4}}{6}} \right) + \frac{3}{6}} \right]^2}\\ = \left( { - \frac{2}{5}} \right) + \frac{{12}}{5}.{\left( {\frac{{ - 1}}{6}} \right)^2}\\ = \left( { - \frac{2}{5}} \right) + \frac{{12}}{5}.\frac{1}{{36}}\\ = \left( { - \frac{2}{5}} \right) + \frac{1}{{15}}\\ = \left( { - \frac{6}{{15}}} \right) + \frac{1}{{15}}\\ = \frac{{ - 5}}{{15}}\\ = \frac{{ - 1}}{3}\end{array}\)             

d)

\(\begin{array}{l}\left\{ {\left[ {{{\left( {\frac{1}{{25}} - 0,6} \right)}^2}:\frac{{49}}{{125}}} \right].\frac{5}{6}} \right\} - \left[ {\left( {\frac{{ - 1}}{3}} \right) + \frac{1}{2}} \right]\\ = \left\{ {\left[ {{{\left( {\frac{1}{{25}} - \frac{3}{5}} \right)}^2}.\frac{{125}}{{49}}} \right].\frac{5}{6}} \right\} - \left[ {\left( {\frac{{ - 2}}{6}} \right) + \frac{3}{6}} \right]\\ = \left\{ {\left[ {{{\left( {\frac{{ 1}}{{25}}-\frac{15}{25}} \right)}^2}.\frac{{125}}{{49}}} \right].\frac{5}{6}} \right\} - \frac{1}{6}\\ = \left\{ {\left[ {{{\left( {\frac{{ - 14}}{{25}}} \right)}^2}.\frac{{125}}{{49}}} \right].\frac{5}{6}} \right\} - \frac{1}{6}\\ = \left\{ {\frac{{196}}{{{{25}^2}}}.\frac{{25.5}}{{49}}.\frac{5}{6}} \right\} - \frac{1}{6}\\ = \left( {\frac{{4.49.25.5.5}}{{{{25}^2}.49.6}}} \right) - \frac{1}{6}\\ = \frac{4}{6} - \frac{1}{6}\\ = \frac{3}{6}\\ = \frac{1}{2}\end{array}\)

27 tháng 1 2019

\(\left(\frac{2}{5}\right)^6.\left(\frac{25}{4}\right)^2\)

\(=\left[\left(\frac{2}{5}\right)^3\right]^2.\left(\frac{25}{4}\right)^2\)

\(=\left[\left(\frac{2}{5}\right)^3.\frac{25}{4}\right]^2\)

\(=\left[\frac{8}{125}.\frac{25}{4}\right]^2\)

\(=\left(\frac{2}{5}\right)^2\)

\(=\frac{4}{25}\)

27 tháng 1 2019

\(15\frac{1}{5}:\left(\frac{-5}{7}\right)-25\frac{1}{5}.\left(\frac{-7}{5}\right)\)

\(=15\frac{1}{5}.\frac{-7}{5}-25\frac{1}{5}.\frac{-7}{5}\)

\(=\frac{-7}{5}\left(15\frac{1}{5}-25\frac{1}{5}\right)\)

\(=\frac{-7}{5}.\left(-10\right)\)

\(=14\)

4 tháng 1 2017

a) \(A=\left(1:\frac{1}{4}\right).4+25\left(1:\frac{16}{9}:\frac{125}{64}\right):\left(-\frac{27}{8}\right)\)

\(=4.4+25.\frac{36}{125}:\frac{-27}{8}\)

\(=16-\frac{32}{15}=\frac{240}{15}-\frac{32}{15}=\frac{208}{15}\)

8 tháng 9 2017

a) \(2^3+3.\left(\frac{1}{2}\right)^0+\left[\left(-2\right)^2:\frac{1}{2}\right]\)

\(=8+3.1+4:\frac{1}{2}\)

\(=8+3+8=19\)

b)\(\frac{2^{15}.9^4}{6^6.8^3}=\frac{2^{15}.\left(3^2\right)^4}{\left(2.3\right)^6.\left(2^3\right)^3}=\frac{2^{15}.3^8}{2^6.3^6.2^9}\)\(=\frac{2^{15}.3^8}{2^{15}.3^6}=3^2=9\)

c) \(\left(1+\frac{2}{3}-\frac{1}{4}\right).\left(\frac{4}{5}-\frac{3}{4}\right)^2\)

\(=\frac{17}{12}.\frac{1}{400}=\frac{17}{4800}\)

d) \(\left(-\frac{10}{3}\right)^3.\left(\frac{-6}{5}\right)^4=-\frac{100}{27}.\frac{1296}{625}\)\(=\frac{-4.48}{1.25}=-\frac{192}{25}\)