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14 tháng 1 2019

3 2   .   1 243 .   81 2 . 1 3 2 = 3 2   .   1 3 5 .   3 4 2 . 1 3 2 = 3 10 3 7 = 3 3 = 27

3 tháng 9 2018

3^2.1/243.81^2.1/3^2

=9/243.81^2/3^2

=1/27.27^2/1

=27^2/27

=27

3 tháng 9 2018

\(3^2\cdot\frac{1}{243}\cdot81^2\cdot\frac{1}{3^2}=\frac{3^2\cdot81^2}{243\cdot3^2}=\frac{3^2\cdot3^8}{3^5\cdot3^2}=3^3=27\)

1 tháng 9 2019

\(A=\left(3+\frac{1}{2}-\frac{2}{3}\right)-\left(2-\frac{2}{3}+\frac{5}{2}\right)+\left(-5+\frac{5}{2}-\frac{4}{3}\right)\)

\(=3+\frac{1}{2}-\frac{2}{3}-2+\frac{2}{3}-\frac{5}{2}-5+\frac{5}{2}-\frac{4}{3}\)

\(=\left(3-2-5\right)+\left(\frac{1}{2}-\frac{5}{2}+\frac{5}{2}\right)-\left(\frac{2}{3}-\frac{2}{3}+\frac{4}{3}\right)\)

\(=-4-\frac{1}{2}\)

\(=-\frac{9}{2}\)

\(B=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{72}+\frac{1}{90}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)

\(=1-\frac{1}{10}\)

\(=\frac{9}{10}\)

1 tháng 9 2019

\(A=\left(3+\frac{1}{2}-\frac{2}{3}\right)-\left(2-\frac{2}{3}+\frac{5}{2}\right)+\left(-5+\frac{5}{2}-\frac{4}{3}\right)\)

\(A=3+\frac{1}{2}-\frac{2}{3}-2+\frac{2}{3}-\frac{5}{2}-5+\frac{5}{2}-\frac{4}{3}\)

\(A=\left(3-2-5\right)+\left(\frac{1}{2}-\frac{5}{2}+\frac{5}{2}\right)-\left(\frac{2}{3}-\frac{2}{3}+\frac{4}{3}\right)\)

\(A=-4+\frac{1}{2}-\frac{4}{3}\)

\(A=-\frac{29}{6}\)

3 tháng 10 2019

\(A=\frac{\frac{1}{3}-\frac{1}{7}-\frac{1}{13}}{\frac{2}{3}-\frac{2}{7}-\frac{2}{13}}.\frac{\frac{1}{3}-0,25+0,2}{\frac{7}{6}-0,875+0,7}+\frac{6}{7}\)

\(=\frac{1}{2}.\frac{2}{7}+\frac{6}{7}=1\)

\(A=\frac{\frac{1}{3}-\frac{1}{7}-\frac{1}{13}}{\frac{2}{3}-\frac{2}{7}-\frac{2}{13}}.\frac{\frac{1}{3}-0,25+0,2}{\frac{7}{6}-0,875+0,7}+\frac{6}{7}\)

\(=\frac{1}{2}-\frac{1}{2}-\frac{1}{2}.\frac{5}{6}-\frac{25}{\frac{100}{-\frac{875}{1000}}}+\frac{6}{7}\)

\(=-\frac{1}{2}.\frac{5}{6}-\frac{25}{\frac{100}{\frac{-87,5}{100}}}+\frac{6}{7}\)

\(=-\frac{1}{2}.\frac{5}{6}-\frac{25}{-87,5}+\frac{6}{7}\)

đến đây tự lm == 

\(=\left(1^3+2^3+3^4+4^5\right)\left(1^3+2^3+3^3+4^3\right)\left(3^8-3^8\right)=0\)

8 tháng 6 2019

\(\left(1^3+2^3+3^4+4^5\right).\left(1^3+2^3+3^3+4^3\right).\left(3^8-81^2\right)\)

\(\left(1^3+2^3+3^4+4^5\right).\left(1^3+2^3+3^3+4^3\right).\left(3^8-3^{4^2}\right)\)

\(\left(1^3+2^3+3^4+4^5\right).\left(1^3+2^3+3^3+4^3\right).\left(3^8-3^8\right)\)

\(\left(1^3+2^3+3^4+4^5\right).\left(1^3+2^3+3^3+4^3\right).0\)

\(=0\)

Chúc bạn học tốt !!!

13 tháng 9 2019

\(B=\frac{\left[\frac{2}{3}\right]^3\cdot\left[-\frac{3}{4}\right]^2\cdot\left[-1\right]^5}{\left[\frac{2}{5}\right]^2\cdot\left[-\frac{5}{12}\right]^3}\)

\(=\frac{\frac{2^3}{3^3}\cdot\frac{\left[-3\right]^2}{4^2}\cdot\left[-1\right]}{\frac{2^2}{5^2}\cdot\frac{\left[-5\right]^3}{12^3}}\)

\(=\frac{\frac{8}{27}\cdot\frac{9}{16}\cdot\left[-1\right]}{\frac{4}{25}\cdot\frac{-125}{\left[2^2\cdot3\right]^3}}\)

\(=\frac{\frac{1}{3}\cdot\frac{1}{2}\cdot\left[-1\right]}{\frac{4}{25}\cdot\frac{-125}{\left[2^2\right]^3\cdot3^3}}\)

\(=\frac{\frac{1\cdot1\cdot\left[-1\right]}{3\cdot2\cdot1}}{\frac{4}{25}\cdot\frac{-125}{4^3\cdot3^3}}\)

\(=\frac{\frac{-1}{6}}{\frac{4}{25}\cdot\frac{-125}{64\cdot27}}=\frac{\frac{-1}{6}}{\frac{4}{1}\cdot\frac{-5}{64\cdot27}}\)

\(=\frac{\frac{-1}{6}}{4\cdot\frac{-5}{64\cdot27}}=\frac{\frac{-1}{6}}{-\frac{20}{64\cdot27}}=\frac{72}{5}\)

9 tháng 9 2019

Gửi tạm trước 2 câu !

\(a,\text{ }3^2\cdot\frac{1}{243}\cdot81^2\cdot3^{-3}=3^2\cdot\frac{1}{3^5}\cdot\left(3^4\right)^2\cdot\frac{1}{3^3}=3^2\cdot\frac{1}{3^5}\cdot3^8\cdot\frac{1}{3^3}=3^2=9\)\(b,\text{ }\frac{\left(-3\right)^{10}\cdot15^5}{25^3\cdot\left(-9\right)^7}=\frac{3^{10}\cdot\left(3\cdot5\right)^5}{\left(5^2\right)^3\cdot\left(-3\cdot3\right)^7}=\frac{3^{10}\cdot3^5\cdot5^5}{5^6\cdot3^7\cdot\left(-3\right)^7}=\frac{3^{15}\cdot5^5}{5^6\cdot3^7\cdot\left(-3\right)^7}=\frac{3}{-5}\)

9 tháng 9 2019

Trả lời :

\(a,\text{ }3^2\cdot\frac{1}{243}\cdot81^2\cdot3^{-3}=3^2\cdot\frac{1}{3^5}\cdot\left(3^4\right)^2\cdot\frac{1}{3^3}=3^2\cdot\frac{1}{3^5}\cdot3^8\cdot\frac{1}{3^3}=3^2=9\)\(b,\text{ }\frac{\left(-3\right)^{10}\cdot15^5}{25^3\cdot\left(-9\right)^7}=\frac{3^{10}\cdot\left(3\cdot5\right)^5}{\left(5^2\right)^3\cdot\left(-3\cdot3\right)^7}=\frac{3^{10}\cdot3^5\cdot5^5}{5^6\cdot3^7\cdot\left(-3\right)^7}=\frac{3^{15}\cdot5^5}{5^6\cdot3^7\cdot\left(-3\right)^7}=\frac{3}{-5}\)

29 tháng 9 2018

a) \(\frac{-2^2}{3}.x=\frac{-2^5}{3}\)

                 \(x=\frac{-2^5}{3}:\frac{-2^2}{3}\)

                 \(x=\frac{-2^5}{3}.\frac{-3}{2^2}\)

                 \(x=\frac{2^3}{1}\)

                 \(x=2^3\)

                 \(x=8\)

Vậy x = 8

b)\(\frac{-1^3}{3}.x=\frac{1}{81}\)

                \(x=\frac{1}{81}:\frac{-1^3}{3}\)

                \(x=\frac{1}{81}.\frac{-3}{1^3}\)

                \(x=\frac{1}{3^4}.\frac{-3}{1^3}\)

                \(x=\frac{1}{3^3}.\frac{-1}{1^2}\)

                \(x=\frac{-1}{3^3}\)

                \(x=\frac{-1}{27}\)

Vậy \(x=\frac{-1}{27}\)

26 tháng 6 2018

a)\(\left(\frac{-1}{3}\right)^3\cdot x=\frac{1}{81}\) \(< =>\frac{-1}{27}x=\frac{1}{81}\)\(< =>x=\frac{-1}{3}\)

28 tháng 9 2018

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