(5.9.3^8)^3
2^3.3^17.9^6-27^10
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Q6
1
15 tháng 8 2023
`@` `\text {Ans}`
`\downarrow`
`a,`
`5.125.25 \div 5^6`
`=`\(5\cdot5^3\cdot5^2\div5^6\)
`=`\(5^{1+3+2-6}=5^{6-6}=5^0=1\)
`b,`
\(2^{14}\div\left(2^6\cdot32\right)\)
`=`\(2^{14}\div\left(2^6\cdot2^5\right)\)
`=`\(2^{14}\div2^{11}=2^3\)
`c,`
`3.3^5\div 27`
`=`\(3\cdot3^5\div3^3\)
`=`\(3^{1+5-3}\)
`=`\(3^3\)
`d,`
\(2^{15}\div\left(2^6\cdot32\right)=2^{15}\div\left(2^6\cdot2^5\right)=2^{15}\div2^{11}=2^4\)
`e,`
\(3^2\cdot27\div81=3^2\cdot3^3\div3^4=3\)
`g,`
\(100\cdot1000\cdot10000-10^9=10^2\cdot10^3\cdot10^4-10^9\)
`=`\(10^9-10^9=0\)
`h,`
\(125^4\div5^9=\left(5^3\right)^4\div5^9=5^{12}\div5^9=5^3\)
MV
29 tháng 10 2017
\(A=\dfrac{12^{15}\cdot3^4-4^5\cdot3^9}{27^3\cdot2^{10}-32^3\cdot3^9}\\ =\dfrac{\left(2^2\cdot3\right)^{15}\cdot3^4-\left(2^2\right)^5\cdot3^9}{\left(3^3\right)^3\cdot2^{10}-\left(2^5\right)^3\cdot3^9}\\ =\dfrac{2^{30}\cdot3^{15}\cdot3^4-2^{10}\cdot3^9}{3^9\cdot2^{10}-2^{15}\cdot3^9}\\ =\dfrac{3^9\cdot2^{10}\left(2^{20}\cdot3^{10}\right)}{3^9\cdot2^{10}\left(1-2^5\right)}\\ =\dfrac{\left(2^2\right)^{10}\cdot3^{10}}{1-32}\\ =\dfrac{\left(2^2\cdot3\right)^{10}}{-31}\\ =\dfrac{-12^{10}}{31}\)
\(B=\dfrac{3}{1^2\cdot2^2}+\dfrac{5}{2^2\cdot3^2}+...+\dfrac{99}{49^2\cdot50^2}\\ =\dfrac{2^2-1^2}{1^2\cdot2^2}+\dfrac{3^2-2^2}{2^2\cdot3^2}+...+\dfrac{50^2-49^2}{49^2\cdot50^2}\\ =\dfrac{1}{1^2}-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+...+\dfrac{1}{49^2}-\dfrac{1}{50^2}\\ =1-\dfrac{1}{2500}\\ =\dfrac{2499}{2500}\)
NN
1
21 tháng 6 2019
\(A=10-\left(\sqrt{32}-\sqrt{8}-\sqrt{27}\right)\left(\sqrt{8}-\sqrt{32}-\sqrt{27}\right)\)
\(A=10-\left[-\sqrt{27}+\left(\sqrt{32}-\sqrt{8}\right)\right]\left[-\sqrt{27}-\left(\sqrt{32}-\sqrt{8}\right)\right]\)
\(A=10-\left[\left(-\sqrt{27}\right)^2-\left(\sqrt{32}-\sqrt{8}\right)^2\right]\)
\(A=10-\left(27-8\right)\)
\(A=-9\)
16 tháng 7 2015
a) 2^7 . 9^3 / 6^5 . 8^2
= 2^7 . 3^9 / 2^5 . 3^5 . 2^6
= 2^-4 . 3^4