\(\frac{\left(\frac{-1}{9}\right)^0.3^2.9^3}{729}\)
Những câu hỏi liên quan
a) \(\frac{1}{2}-\frac{1}{4}.\left(-\frac{6}{5}\right)\)
b) \(\frac{\left(\frac{1}{9}\right)^0.3^2.9^3}{729}\)
a) \(\frac{1}{2}-\frac{1}{4}.\left(-\frac{6}{5}\right)\)
\(=\frac{1}{2}-\left(-\frac{3}{10}\right)\)
\(=\frac{4}{5}\)
b) \(\frac{\left(\frac{1}{9}\right)^0.3^2.9^3}{729}=\frac{1.9.729}{729}=\frac{9.729}{729}=9\)
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\(\dfrac{\left(\dfrac{1}{9}\right)^0.3^2.9^3}{729}\)
A = ( \(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\) ) \(\frac{729}{364}+100:\subset5X\left(3-1\right)\)
1.Tính nhanh
a)427-98
b)2*19*15+3*43*10+62*80
c)\(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
d)\(\left(1-\frac{1}{9}\right)\cdot\left(1-\frac{2}{9}\right)\cdot\left(1-\frac{3}{90}\right)\cdot.........\cdot\left(1-\frac{2018}{9}\right)\)
\(a)\) \(427-98=329\)
\(b)\) \(2\cdot19\cdot15+3\cdot43\cdot10+62\cdot80\)
\(=\left(2\cdot15\right)\cdot19+\left(3\cdot10\right)\cdot43+62\cdot80\)
\(=30\cdot19+30\cdot43+62\cdot80\)
\(=30\cdot\left(19+43\right)+62\cdot80\)
\(=30\cdot62+62\cdot80\)
\(=62\cdot\left(30+80\right)\)
\(=62\cdot110=6820\)
\(c)\) 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 \(M=\frac{364}{729}\)
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cho biểu thức:A=[6\(\times\left(-\frac{1}{3}\right)^2-3\times\left(-\frac{1}{3}\right)+1\)]\(\div\left(-\frac{1}{3}-1\right)\)
B=\(\left(729-1^3\right)\times\left(729-3^3\right)\times...\times\left(729-125^3\right)\)
hãy so sánh A và B
nhanh lên mai mình phải nộp rồi
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
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tính:\(\frac{\left(-12\right)^5.27^4-32^2.81^4}{729^4:\left(-9\right)^4.16^5:\left(-8\right)^3}\)
tính:\(\frac{\left(-12\right)^5.27^4-32^2.81^4}{729^4:\left(-9\right)^4.16^5:\left(-8\right)^3}\)
P/S : Good Luck
~Best Best~
\(\dfrac{1}{2}\)-\(\dfrac{3}{4}\).\(\left(\dfrac{-6}{5}\right)\)
\(\dfrac{\dfrac{1^0}{9}.3^2.9^3}{729}\)
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) \(\dfrac{\dfrac{1^0}{9}.3^2.9^3}{729}\)
\(=\dfrac{9^{-1}.3^2.9^3}{729}\)
\(=\dfrac{9^{-1}.9.9^3}{729}\)
\(=\dfrac{9^{-1+1+3}}{729}\)
\(=\dfrac{9^3}{729}\)
\(=\dfrac{729}{729}\)
\(=1\)
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tính giá trị biểu thức
\(\frac{\frac{1}{4}.3^2.9^3}{729}\)
giúp mik với mik đang cần gấp