\(\frac{\left(\frac{-1}{9}\right)^0.3^2.9^3}{729}\)
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13 tháng 12 2016
a) \(\frac{1}{2}-\frac{1}{4}.\left(-\frac{6}{5}\right)\)
\(=\frac{1}{2}-\left(-\frac{3}{10}\right)\)
\(=\frac{4}{5}\)
b) \(\frac{\left(\frac{1}{9}\right)^0.3^2.9^3}{729}=\frac{1.9.729}{729}=\frac{9.729}{729}=9\)
MT
4 tháng 8 2018
\(a)\) \(427-98=329\)
\(b)\) \(2\cdot19\cdot15+3\cdot43\cdot10+62\cdot80\)
\(=\left(2\cdot15\right)\cdot19+\left(3\cdot10\right)\cdot43+62\cdot80\)
\(=30\cdot19+30\cdot43+62\cdot80\)
\(=30\cdot\left(19+43\right)+62\cdot80\)
\(=30\cdot62+62\cdot80\)
\(=62\cdot\left(30+80\right)\)
\(=62\cdot110=6820\)
\(c)\) Đặt \(M=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}+\frac{1}{3^6}\)
\(\Rightarrow3M=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\)
\(\Rightarrow3M-M=\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}+\frac{1}{3^6}\right)\)
\(\Rightarrow2M=1-\frac{1}{3^6}\)
\(\Rightarrow M=\frac{728}{2\cdot729}=\frac{364}{729}\)
Vậy \(M=\frac{364}{729}\)
26 tháng 2 2017
Ta có: \(A=\left[6.\left(-\frac{1}{3}\right)^2-3.\left(-\frac{1}{3}\right)+1\right]:\left(\frac{-1}{3}-1\right)\)
\(=\left(6.\frac{1}{9}-\left(-1\right)+1\right):\left(\frac{-4}{3}\right)\)
\(=\left(\frac{2}{3}+2\right).\left(\frac{-3}{4}\right)\)
\(=\frac{8}{3}.\left(-\frac{3}{4}\right)\)
\(=-2\)
\(B=\left(729-1^3\right)\left(729-3^3\right)...\left(729-125^3\right)\)
\(\Rightarrow B=\left(729-1^3\right)\left(729-3^3\right)...\left(729-9^3\right)...\left(729-125^3\right)\)
\(\Rightarrow B=\left(729-1^3\right)\left(729-3^3\right)...0...\left(729-125^3\right)\)
\(\Rightarrow B=0\)
Vì -2 < 0 nên A < B
Vậy A < B
NT
0
NN
6 tháng 12 2017
a) \(\dfrac{1}{2}-\dfrac{3}{4}.\dfrac{-6}{5}\)
\(=\dfrac{1}{2}-\dfrac{3.\left(-6\right)}{4.5}\)
\(=\dfrac{1}{2}-\dfrac{-18}{20}\)
\(=\dfrac{1}{2}+\dfrac{9}{10}\)
\(=\dfrac{5}{10}+\dfrac{9}{10}\)
\(=\dfrac{5+9}{10}\)
\(=\dfrac{14}{10}\)
\(=\dfrac{7}{5}\)
b) \(\dfrac{\dfrac{1^0}{9}.3^2.9^3}{729}\)
\(=\dfrac{9^{-1}.3^2.9^3}{729}\)
\(=\dfrac{9^{-1}.9.9^3}{729}\)
\(=\dfrac{9^{-1+1+3}}{729}\)
\(=\dfrac{9^3}{729}\)
\(=\dfrac{729}{729}\)
\(=1\)
HD
0
Trả lời
(-1/9)0.32.93/729
=1.32.93/93
=32=9
\(\frac{\left(-\frac{1}{9}\right)^0.3^2.3^3}{729}=\frac{1.3^5}{3^6}=\frac{1}{3}\)