3^17.81^11/27^10.9^15
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6 tháng 7 2017
a)\(\frac{15\times3^5}{27^3\times10^2}=\frac{5\times3^6}{3^9\times5\times20}=\frac{1}{3^3\times20}=\frac{1}{540}\)
b) đề thiếu
c) \(\frac{81^{14}\times3^{17}}{27^{10}\times9^{15}}=\frac{\left(9^2\right)^{14}\times3^{17}}{3^{10}\times9^{10}\times9^{15}}=\frac{9^{28}\times3^{17}}{3^{10}\times9^{25}}=\frac{9^3\times3^7}{1}=1594323\)
7 tháng 6 2016
a) =\(\left(\frac{1}{4}\right)^3\cdot2^5=\frac{1}{2^6}2^5=0,5\)
b) \(=\left(\frac{1}{8}\right)^3\cdot8^4\cdot10^4=80.000\)
c) \(=8^2\cdot\frac{4^5}{2^{20}}=\frac{2^6\cdot2^{10}}{2^{20}}=\frac{1}{2^4}=0,0625\)
d) = \(\frac{81^{11}\cdot3^{17}}{27^{10}\cdot9^{15}}=\frac{3^{44}\cdot3^{17}}{3^{30}\cdot3^{30}}=\frac{3^{61}}{3^{60}}=3\)
16 tháng 4 2016
a) 81^11.3^17/27^10.9^15
=(9^2)^11.3^17/(3^3)^10.9^15
=3^44.3^17/3^30.3^30
=3^61/3^60
=3
b) A =( (2^12.3^5 - 2^12.3^4)/ (2^12.3^6 + 2^12.3^5) ) - ((5^10.7^3 - 5^10.7^4)/(5^9.7^3 + 5^9.2^3.7^3))
=(2^12.3^4(3-1))/2^12.3^5(3+1) - 5^10.7^3(1-7)/5^9.7^3(1+8)
=2/12- (-30/9)=1/6 + 10/3 = 7/2
NB
0
2 tháng 6 2022
Bài 2:
a: \(\left(0.25\right)^3\cdot32=\dfrac{1}{4^3}\cdot32=\dfrac{32}{64}=\dfrac{1}{2}\)
b: \(\left(-0.125\right)^3\cdot80^4=\left(-0.125\cdot80\right)^3\cdot80=-80\)
c: \(\dfrac{8^2\cdot4^5}{2^{20}}=\dfrac{2^6\cdot2^{10}}{2^{20}}=\dfrac{1}{2^4}=\dfrac{1}{16}\)
d: \(\dfrac{81^{11}\cdot3^{17}}{27^{10}\cdot9^{15}}=\dfrac{3^{44}\cdot3^{17}}{3^{30}\cdot3^{30}}=\dfrac{3^{61}}{3^{60}}=3\)
19 tháng 7 2018
\(\frac{5-\frac{5}{3}+\frac{5}{9}-\frac{5}{27}}{8-\frac{8}{3}+\frac{8}{9}-\frac{8}{27}}:\frac{15-\frac{15}{11}+\frac{15}{121}}{16-\frac{16}{11}+\frac{16}{121}}\)
\(=\frac{5\left(1-\frac{1}{3}+\frac{1}{9}-\frac{1}{27}\right)}{8\left(1-\frac{1}{3}+\frac{1}{9}-\frac{1}{27}\right)}:\frac{15\left(1-\frac{1}{11}+\frac{1}{121}\right)}{16\left(1-\frac{1}{11}+\frac{1}{121}\right)}\)
\(=\frac{5}{8}:\frac{15}{16}\)
\(=\frac{2}{3}\)
[3^17.81^11]/[27^10.9^15]
=[3^17.(3^4)^11]/[(3^3)^10.(3^2)^15]
=[3^17.3^44]/[3^30.3^30]
=3^61/3^60
=3
\(\frac{3^{17}.81^{11}}{27^{10}.9^{15}}\)
\(=\)\(\frac{3^{17}.\left(3^4\right)^{11}}{\left(3^3\right)^{10}.\left(3^2\right)^{15}}\)
\(=\)\(\frac{3^{17}.3^{44}}{3^{30}.3^{30}}\)
\(=\)\(\frac{3^{61}}{3^{60}}\)
\(=\)\(3\)