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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) = \(\frac{81^{11}\cdot3^{17}}{27^{10}\cdot9^{15}}=\frac{3^{44}\cdot3^{17}}{3^{30}\cdot3^{30}}=\frac{3^{61}}{3^{60}}=3\)
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: \(\dfrac{81^{11}\cdot3^{17}}{27^{10}\cdot9^{15}}=\dfrac{3^{44}\cdot3^{17}}{3^{30}\cdot3^{30}}=\dfrac{3^{61}}{3^{60}}=3\)
\(=\dfrac{1}{2}-\left(\dfrac{1}{3\cdot7}+\dfrac{1}{7\cdot11}+...+\dfrac{1}{23\cdot27}\right)\)
\(=\dfrac{1}{2}-\dfrac{1}{4}\left(\dfrac{4}{3\cdot7}+\dfrac{4}{7\cdot11}+...+\dfrac{4}{23\cdot27}\right)\)
\(=\dfrac{1}{2}-\dfrac{1}{4}\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{23}-\dfrac{1}{27}\right)\)
\(=\dfrac{1}{2}-\dfrac{1}{4}\cdot\dfrac{9-1}{27}\)
\(=\dfrac{1}{2}-\dfrac{1}{4}\cdot\dfrac{8}{27}=\dfrac{1}{2}-\dfrac{2}{27}=\dfrac{27-4}{54}=\dfrac{23}{54}\)
\(\dfrac{3^{17}.81^{11}}{27^{10}.9^{15}}=\dfrac{3^{17}.\left(3.27\right)^{11}}{27^{10}.\left(3^2\right)^{15}}=\dfrac{3^{17}.3^{11}.27^{11}}{27^{10}.3^{30}}=\dfrac{3^{28}.27^{10}.27}{27^{10}.3^{28}.3^2}=^{ }\dfrac{27}{3^2}=\dfrac{27}{9}=3\)
Do x2n=(-x)2n
=>327.1530.(-4)16=327.(-15)30.416=327.(-15)30.(42)8=327.(-15)30.88
(327.1530.(-4)16)/((-15)30.811)=(327.(-15)30.88)/((-15)30.811)=327/83
[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
\(\frac{3^{17}.81^{11}}{27^{10}.9^{15}}\)
\(=\)\(\frac{3^{17}.\left(3^4\right)^{11}}{\left(3^3\right)^{10}.\left(3^2\right)^{15}}\)
\(=\)\(\frac{3^{17}.3^{44}}{3^{30}.3^{30}}\)
\(=\)\(\frac{3^{61}}{3^{60}}\)
\(=\)\(3\)