Baì 1 Chứng minh rằng
a)\(7^6+7^5-7^4⋮11\)
b)\(81^7-27^9-9^{13}⋮45\)
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a/ \(5^5-5^4+5^3=5^3\left(5^2-5+1\right)=5^3.21⋮7\left(đpcm\right)\)
b/ \(7^6+7^5-7^4=7^4\left(7^2+7-1\right)=7^4.55⋮11\left(đpcm\right)\)
c/ \(10^9+10^8+10^7=10^7.\left(10^2+10+1\right)=10^7.111=1110000⋮222\left(đpcm\right)\)
d/ \(10^6-5^7=2^6.5^6-5^7=5^6\left(2^6-5\right)=5^6.59\left(đpcm\right)\)
e/ \(3^{n+2}-2^{n+2}+3^n-2^n=3^n\left(3^2+1\right)-2^n\left(2^2+1\right)=3^n.10-2^n.5=3^n.10-2^{n-1}.10=10\left(3^n-2^{n-1}\right)⋮10\left(đpcm\right)\)
f/ \(81^7-27^9-9^{13}=3^{28}-3^{27}-3^{26}=3^{26}\left(3^2-3-1\right)=3^{26}.5=3^{24}.45⋮45\left(đpcm\right)\)
a) Ta có: 55 - 54 + 53
= 53(52 - 5 + 1)
= 53 . 3 . 7 \(⋮\) 7 (đpcm)
Bài 2:
Ta có: \(\frac{\left(3^3\right)^2.\left(2^3\right)^5}{\left(2.3\right)^6.\left(2^5\right)^3}\)\(=\frac{3^6.2^{15}}{2^6.3^6.2^{15}}\)\(\frac{1}{2^6}=\frac{1}{64}\)
Chúc hk tốt nha!!!
Lời giải:
a) Ta có:
\(7^6+7^5-7^4=7^{4+2}+7^{4+1}-7^4\)
\(=7^4.7^2+7^4.7-7^4=7^4(7^2+7-1)=7^4.55=11.7^4.5\vdots 11\) (đpcm)
b)
\(81^7-27^9-9^{13}=(3^4)^7-(3^3)^9-(3^2)^{13}\)
\(=3^{28}-3^{27}-3^{26}\)
\(=3^{26}(3^2-3-1)=5.3^{26}=5.3.3.3^{24}=45.3^{24}\vdots 45\) (đpcm)
a, \(7^6+7^5-7^4⋮11\)
= \(7^4.7^2+7^4.7-7^4\)
= \(7^4.\left(7^2+7-1\right)\)
= \(7^4.\left(49+7-1\right)\)
=\(7^4.55=7^4.5.11\) => chia hết cho 11
b, \(81^7\)- \(27^9\)- \(9^{13}\)
=\(\left(3^4\right)^7\)- \(\left(3^3\right)^9\) - \(\left(3^2\right)^{13}\)
= \(3^{28}-3^{27}-3^{26}\)
=\(3^{26}.\left(3^2-3-1\right)\)
=3^26.5=3^13.3^2.5=45.3^13 chia hết cho 45
a)
\(7^6+7^5-7^4=7^4\left(7^2+7-1\right)=7^4.55\) chia hết cho 55 (đpcm )
b)
\(16^5+2^{15}=\left(2^4\right)^5+2^{15}=2^{20}+2^{15}=2^{15}\left(2^5+1\right)=2^{15}.33\) chia hết cho 33 (đpcm )
c)
\(81^7-27^9-9^{13}=\left(3^4\right)^7-\left(3^3\right)^9-\left(3^2\right)^{13}=3^{28}-3^{27}-3^{26}\)
\(=3^{22}\left(3^6-3^5-3^4\right)=3^{22}.405\) chia hết cho 405 (đpcm )
\(=\left(3^4\right)^7-\left(3^3\right)^9-\left(3^2\right)^{13}=3^{28}-3^{27}-3^{26}=\)
\(=3^{26}\left(3^2-3-1\right)=3^{26}.5⋮5\)
Nguyễn Ngọc Quý sai ...= 7^6. ( 7-1+49)= 7^6.55 chia hết cho 11
7^6-7^5+7^9=7^5nhân(7-1+7^4)=7^5nhân 55=vì 55 chia hết cho 11,nên7^6-7^5+7^9 chia hết cho11
Bài 1:
a) \(7^6+7^5-7^4\)
\(=7^4.7^2+7^4.7-7^4\)
\(=7^4.\left(7^2+7-1\right)\)
\(=7^4.\left(49+7-1\right)\)
\(=7^4.55\)
\(=7^4.5.11\)
Vì \(11⋮11\) nên \(7^4.5.11⋮11\)
\(\Rightarrow7^6+7^5-7^4⋮11.\)
b) \(81^7-27^9-9^{13}\)
\(=\left(3^4\right)^7-\left(3^3\right)^9-\left(3^2\right)^{13}\)
\(=3^{28}-3^{27}-3^{26}\)
\(=3^{26}.\left(3^2-3-1\right)\)
\(=3^{26}.5\)
\(=3^{13}.3^2.5\)
\(=3^{13}.45\)
Vì \(45⋮45\) nên \(3^{13}.45⋮45\)
\(\Rightarrow81^7-27^9-9^{13}⋮45.\)
Chúc bạn học tốt!
a) 7^6+7^5-7^4 chia hết cho 11
= 7^4 ( 7^2 + 7 - 1 )
= 7^4 ( 49 +7 - 1 )
= 7^4 + 55
= 7^4 x 5 x 11 chia hết cho 11 ( đpcm )
b ) 81^7-27^9-9^13 chia hết cho 45
= (3^4)^7 - ( 3^3)^9 - ( 3^2 )^13
= 3^ 28- 3 ^27 - 3^26
= 3 ^26 x ( 3^2 - 3^1 - 3^0 )
= 3^24 x 9 x5
= 3 ^24 x 45 chia hết cho 45 ( đpcm )