\(\left(5^2+1\right).\left(5^4+1\right).\left(5^8+1\right).\left(5^{16...">
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30 tháng 7 2018

\(P=12.\left(5^2+1\right).\left(5^4+1\right).\left(5^8+1\right).\left(5^{16}+1\right)\)

\(=\frac{1}{2}\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\frac{1}{2}\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\frac{1}{2}\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\frac{1}{2}\left(5^{16}-1\right)\left(5^{16}+1\right)\)

\(=\frac{5^{32}-1}{2}\)

6 tháng 11 2017

a, \(\dfrac{2\cdot8^4\cdot27^2+4\cdot6^9}{2^7\cdot6^7+2^7\cdot40\cdot9^4}\)

=\(\dfrac{2\cdot\left(2^3\right)^4\cdot\left(3^3\right)^2+2^2\cdot2^9\cdot3^9}{2^7\cdot2^7\cdot3^7+2^7\cdot2^3\cdot5\cdot\left(3^2\right)^4}\)

=\(\dfrac{2\cdot2^{12}\cdot3^6+2^{11}\cdot3^9}{2^{14}\cdot3^7+2^{10}\cdot5\cdot3^8}\)

=\(\dfrac{2^{11}\cdot3^6\cdot\left(2^2+3^3\right)}{2^{10}\cdot3^7\cdot\left(2^4+5\cdot3\right)}\)

=\(\dfrac{2^{11}\cdot3^6\cdot31}{2^{10}\cdot3^7\cdot31}\)

=\(\dfrac{2}{3}\)

6 tháng 11 2017

b, \(\dfrac{\dfrac{8}{27}\cdot\dfrac{9}{16}\cdot\left(-1\right)}{\dfrac{4}{25}\cdot\dfrac{-125}{1728}}\)

=\(\dfrac{\dfrac{8\cdot9\cdot\left(-1\right)}{27\cdot16}}{\dfrac{4\cdot\left(-125\right)}{25\cdot1728}}\)

=\(\dfrac{\dfrac{-1}{6}}{\dfrac{-5}{432}}\)

=\(\dfrac{-1}{6}\cdot\dfrac{-432}{5}\)

=\(\dfrac{72}{5}\)

14 tháng 3 2019

KQ:\(\frac{1}{5}\)

14 tháng 3 2019

cho tớ xin cách lm

4 tháng 10 2021

yutyugubhujyikiu

17 tháng 7 2016

Cũng khuya rồi , mình làm câu 1 thôi nhé !
\(\frac{2.5^{22}-9.5^{21}}{25^{10}}=\frac{2.5^{22}-9.5^{21}}{\left(5^2\right)^{10}}\)

\(\frac{5^{21}.\left(2.5-9\right)}{5^{20}}=5.\left(10-9\right)=5\)
 

18 tháng 7 2017

\(\dfrac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\dfrac{2^{10}.3^8-2.3^9.2^9}{2^{10}.3^8+2^8.3^8.2^2.5}=\dfrac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}\)

\(=\dfrac{2^{10}.\left(3^8-3^9\right)}{2^{10}.3^8.\left(1+5\right)}=\dfrac{3^8-3^9}{3^8.6}=\dfrac{3^8.\left(1-3\right)}{3^8.6}=\dfrac{-2}{6}=-\dfrac{1}{3}\)

~ Học tốt ~

18 tháng 7 2017

Bài 1:

1) \(3^2.\dfrac{1}{243}.81^2.\dfrac{1}{3^3}\)

\(=3^2.\left(\dfrac{1}{3}\right)^5.\left(3^4\right)^2.\dfrac{1}{3^3}\)

\(=3^2.\dfrac{1}{3^5}.3^8.\dfrac{1}{3^3}\)

\(=3^2=9\)

2) \(\left(4.2^5\right):\left(2^3.\dfrac{1}{16}\right)\)

\(=\left(2^2.2^5\right):[2^3.\left(\dfrac{1}{2}\right)^4]\)

\(=2^7:2^3:\dfrac{1}{2^4}\)

\(=2^4.2^4=256\)

3)\(\left(2^{-1}+3^{-1}\right)+\left(2^{-1}.2^0\right):2^3\)

\(=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{2}.1:2^3\)

\(=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{2^4}\)

\(=\dfrac{43}{48}\)

4)\(\left(-\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2:2\)

\(=-3-1+\dfrac{1}{4}.\dfrac{1}{2}\)

\(=-3-1+\dfrac{1}{8}\)

\(=-4+\dfrac{1}{8}\\ \)

\(=-\dfrac{31}{8}\)

5)\([\left(0,1\right)^2]^0+[\left(\dfrac{1}{7}\right)^{-1}]^2.\dfrac{1}{49}.[\left(2^2\right)^3:2^5]\\ =1+7^2.\dfrac{1}{7^2}.2^6:2^5\\ =1+1.2\\ =3\)

Chúc bạn học tốt haha