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MC
3 tháng 8 2018
\(1.a)\) Ta có: \(\left\{{}\begin{matrix}64^8=\left(8^2\right)^8=8^{16}\\16^{12}=8^{12}.2^{12}=8^{12}.\left(2^3\right)^4=8^{12}.8^4=8^{16}\end{matrix}\right.\)
Có: \(8^{16}=8^{16}\Rightarrow64^8=16^{12}\)
Vậy...
\(b)\) Ta có: \(\left\{{}\begin{matrix}\left(-5\right)^{30}=\left[\left(-5\right)^3\right]^{10}=\left(-125\right)^{10}\\\left(-3\right)^{50}=\left[\left(-3\right)^5\right]^{10}=\left(-243\right)^{10}\end{matrix}\right.\)
Có: \(\left(-125\right)^{10}< \left(-243\right)^{10}\Rightarrow\left(-5\right)^{30}< \left(-3\right)^{50}\)
Vậy...
\(c)\) Ta có: \(\left\{{}\begin{matrix}2^{27}=\left(2^3\right)^9=8^9\\3^{18}=\left(3^2\right)^9=9^9\end{matrix}\right.\)
Có: \(8^9< 9^9\Rightarrow2^{27}< 3^{18}\)
Vậy...
\(d)\) Ta có: \(\left\{{}\begin{matrix}\left(\dfrac{1}{25}\right)^{10}=\left[\left(\dfrac{1}{5}\right)^2\right]^{10}=\left(\dfrac{1}{5}\right)^{20}\\\left(\dfrac{1}{125}\right)^8=\left[\left(\dfrac{1}{5}\right)^3\right]^8=\left(\dfrac{1}{5}\right)^{24}\end{matrix}\right.\)
Có: \(\left(\dfrac{1}{5}\right)^{20}< \left(\dfrac{1}{5}\right)^{24}\Rightarrow\left(\dfrac{1}{24}\right)^{10}< \left(\dfrac{1}{125}\right)^8\)
Vậy...
\(e)\)Có: \(32^9=\left(2^5\right)^9=2^{45}< 2^{52}=\left(2^4\right)^{13}=16^{13}< 18^{13}\)
\(\Rightarrow32^9< 18^{13}\)
Vậy...
HH
1
15 tháng 8 2017
a, \(81^7-27^9-9^{13}\)
\(=\left(3^4\right)^7-\left(3^3\right)^9-\left(3^2\right)^{13}\)
\(=3^{28}-3^{27}-3^{26}\)
\(=3^{26}\left(3^2-3-1\right)\)
\(=3^{25}.3.5\)
\(=3^{25}.15⋮15\)
\(\Leftrightarrow81^7-27^9-9^{13}⋮15\Leftrightarrowđpcm\)
HH
1
15 tháng 8 2017
a. Ta có: \(81^7-27^9-9^{13}\)
\(=\left(3^4\right)^7-\left(3^3\right)^9-\left(3^2\right)^{13}=3^{28}-3^{27}-3^{26}\)
\(=3^{25}\left(3^3-3^2-3\right)=3^{25}\left(27-9-3\right)=3^{25}\cdot15\)
Vì \(15⋮15\) nên \(3^{25}\cdot15⋮15\)
\(\Rightarrow81^7-27^9-9^{13}⋮15\) (đpcm)
b. Ta có: \(24^{54}\cdot54^{24}\cdot2^{10}\)
\(=\left(2^3\cdot3\right)^{54}\cdot\left(3^3\cdot2\right)^{24}\cdot2^{10}\)
\(=\left(2^3\right)^{54}\cdot3^{54}\cdot\left(3^3\right)^{54}\cdot2^{54}\cdot2^{10}\)
\(=2^{162}\cdot2^{24}\cdot2^{10}\cdot3^{54}\cdot3^{72}=2^{196}\cdot3^{126}\)
Mà \(72^{63}=\left(2^3\cdot3^2\right)^{63}\)
\(=\left(2^3\right)^{63}\cdot\left(3^2\right)^{63}=2^{189}\cdot3^{126}\)
Vì \((2^{196}\cdot3^{126})⋮\left(2^{189}\cdot3^{126}\right)\)
\(\Rightarrow24^{54}\cdot54^{24}\cdot2^{10}⋮72^{63}\) (đpcm)
7 tháng 7 2016
81^7-27^9-9^13
=(3^4)^7-(3^3)^9-(3^2)^13
=3^28-3^27-3^26
=(3^26.3^2)-(3^26.3^1)-(3^26.1)
=3^26.(9-3-1)
=3^22.(3^4.5)
=3^22.405 chia het cho 405
=> 81^7-27^9-9^13 chia het cho 405
7 tháng 7 2016
\(81^7-27^9-9^{13}=\left(3^4\right)^7-\left(3^3\right)^9-\left(3^2\right)^{13}\)
\(=3^{28}-3^{27}-3^{36}=3^{22}.\left(3^6-3^5-3^4\right)\)
\(=3^{22}.405⋮405\)
9 tháng 8 2017
a) 8110 - 2713 - 921 \(⋮\)225
= 8110 - 273 - 921 : 225
= ( 34 )10 - ( 33 )3 - ( 32 )21
= 340 - 39 - 342
= 339 . ( -25 )
= -225 . 337
-225 nhân 1 số tự nhiên thì luôn luôn chia hết cho 225
8 tháng 7 2021
+)\(8^2=\left(2^3\right)^2=2^6\)
+)\(3^{200}=3^{2.100}=\left(3^2\right)^{100}=9^{100}\)
\(2^{300}=2^{3.100}=\left(2^3\right)^{100}=8^{100}\)
Vì \(9>8\Rightarrow9^{100}>8^{100}\)hay \(3^{200}>2^{300}\)
+)\(9^{20}=\left(3^2\right)^{20}=3^{40}\)
\(27^{13}=\left(3^3\right)^{13}=3^{39}\)
Vì \(40>39\Rightarrow3^{40}>3^{39}\)hay \(9^{20}>27^{13}\)
+)\(10^{20}=10^{2.10}=\left(10^2\right)^{10}=100^{10}\)
\(2^{100}=2^{10.10}=\left(2^{10}\right)^{10}=1024^{10}\)
Vì \(100< 1024\Rightarrow100^{10}< 1024^{10}\)hay \(10^{20}< 2^{100}\)
+)\(2^{161}=2^{4.40+1}=\left(2^4\right)^{40}.2=16^{40}.2\)
Vì \(13< 16\Rightarrow13^{40}< 16^{40}\)\(\Rightarrow13^{40}< 2^{161}\)
DQ
1
D
3 tháng 11 2017
a) Ta có: \(81^7-27^9-9^{13}=\left(3^4\right)^7-\left(3^3\right)^9-\left(3^2\right)^{13}\)
\(=3^{28}-3^{27}-3^{26}=3^{22}\left(3^6-3^5-3^4\right)\)
\(=3^{22}\times405\)
\(\Rightarrow81^7-27^9-9^{13}⋮405\)(vì có chứa thừa số 405)
b) Ta có: \(8^7-2^{18}=\left(2^3\right)^7-2^{18}=2^{21}-2^{18}\)
\(=2^{17}\left(2^4-2\right)=2^{17}\times14\)
\(\Rightarrow8^7-2^{18}⋮14\)(vì có chứa thừa số 14)
HT
1
22 tháng 5 2022
\(A=\dfrac{3^6\cdot3^8\cdot5^4-5^{13-9}\cdot3^{13}}{3^{12}\cdot5^6+3^{12}\cdot5^6}\)
\(=\dfrac{3^{14}\cdot5^4-5^4\cdot3^{13}}{3^{12}\cdot5^6\cdot2}\)
\(=\dfrac{3^{13}\cdot5^4\cdot2}{3^{12}\cdot5^6\cdot2}=\dfrac{3}{25}\)
\(B=\left(\dfrac{2}{5}\cdot5\right)^7+\left(\dfrac{9}{4}:\dfrac{3}{16}\right)^3=1+12^3=1729\)
\(27^{13}\times81^{12}\)
\(=\left(3^3\right)^{13}\times\left(3^4\right)^{12}\)
\(=3^{39}\times3^{48}\)
\(=3^{39+48}\)
\(=3^{87}\)