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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)
a, \(81^7-27^9-9^{13}\)
\(=\left(3^4\right)^7-\left(3^3\right)^9-\left(3^2\right)^{13}\)
\(=3^{28}-3^{27}-3^{26}\)
\(=3^{26}\left(3^2-3-1\right)\)
\(=3^{25}.3.5\)
\(=3^{25}.15⋮15\)
\(\Leftrightarrow81^7-27^9-9^{13}⋮15\Leftrightarrowđpcm\)
tham khảo câu b bài 1 ở link này https://olm.vn/hoi-dap/detail/88152567739.html
\(5^5-5^4+5^3=5^3.5^2-5^3.5+5^3=5^3.(5^2-5+1)\)
\(=5^3.21=5^3.3.7 \vdots 7 \Rightarrow 5^5-5^4+5^3\vdots 7\)
Tương tự :
b,\(7^6+7^5-7^4=7^4.(7^2+7-1)=7^4.55=7^4.5.11\vdots11\)
\(\Rightarrow 7^6+7^5-7^4\vdots 11\)
c,\(24^{54}.54^{24}.2^{10}=(2^3.3)^{54}.(2.3^3)^{24}.2^{10}\)
\(=(2^3)^{54}.3^{54}.2^{24}.(3^3)^{24}.2^{10}\)
\(=(2^3)^{54}.(2^3)^8.2^3.(3^2)^{27}.(3^2)^{36}.2^{7}\)
\(=(2^3)^{63}.(3^2)^{63}.2^7=(2^3.3^2)^{63}.2^7=72^{63}.2^7 \vdots 72^{63}\)
d,\(3^{n+3}+3^{n+1}+2^{n+3}.2^{n+2}=3^{n+1}.3^2+3^{n+1}+2^{n+3}.2^{n+2}\)
\(=3^{n+1}.(3^2+1)+2^{2n+5}=10.3^{n+1}+2.2^{2n+4}\)
\(=2.(5.3^{n+1}+2^{2n+4})\)
Lỗi đề rồi!!!!!!!!!! tớ thay số vào không đúng!
a) \(7^6+7^5-7^4\)chia hết cho 11
\(=7^4\left(7^2+7-1\right)\)
\(=7^4.55=7^4.5.11\)chia hết cho 11
b) \(24^{54}.54^{24}.2^{10}\)chia hết cho \(72^{63}\)
\(=\left(2^3.3\right)^{54}.\left(3^3.2\right)^{24}\)
\(=\left(2^3\right)^{54}.3^{54}.\left(3^3\right)^{24}.2^{24}.2^{10}\)
\(=2^{162}.2^{24}.2^{10}.3^{54}.3^{72}\)
\(=2^{196}.3^{126}\)
\(72^{63}=\left(2^3.3^2\right)^{63}\)
\(=\left(2^3\right)^{63}.\left(3^2\right)^{63}=2^{189}.3^{126}\)
Vì \(2^{196}.3^{126}\)chia hết \(2^{189}.3^{126}\)
\(\Rightarrow24^{54}.54^{24}.2^{10}\)chia hết cho\(72^{63}\)
Câu a:
A = (36\(^{36}\) - 9\(^{10}\))
A = [\(\overline{..6}\) - (9\(^2\))\(^{10}\) ]
A = [\(\overline{..6}\) - (\(\overline{..1}\))\(^{10}\)]
A = [\(\overline{..6}\) - \(\overline{..1}\)]
A = \(\overline{..5}\)
Vậy A ⋮ 5 (1)
A = (36\(^{36}\) - 9\(^{10}\))
Vì 36 ⋮ 9; 9 ⋮ 9 nên A ⋮ 9 (2)
Từ (1) và (2) ta có:
A ⋮ 5 và 9
5 = 5; 9 = 3\(^2\)
BCNN(5; 9) = 5.3\(^2\) = 45 suy ra A ∈ B(45) hay A ⋮ 45(đpcm)
b) B = [12^(n+1)+11^(n+2)] ⋮ 133 ∀n nguyên dương
Giải:
Với n = 1 ta có:
B = [12\(^{1+1}\) + 11\(^{1+2}\)]
B = [12\(^2\) + 11\(^3\)]
B = [144 + 1331]
B = 1475
1475 : 133 = 11 dư 12
Vậy việc chứng minh B chia hết cho 133 với mọi n nguyên dương là không thể.
Bạn tham khảo nhé! Mình không chắc là đúng hay không nữa 
https://hoc24.vn/hoi-dap/question/124418.html
\(24^{54}.54^{24}.2^{10}=\left(2^3.3\right)^{54}.\left(3^3.2\right)^{24}.2^{10}\)\(=2^{3.54}.3^{54}.3^{3.24}.2^{24}.2^{10}\)
\(=2^{162}.3^{54}.3^{72}.2^{72}.2^{24}.2^{10}=2^{162+72+54+10}.3^{54+72}\)\(=2^{298}.3^{126}\).
\(72=3^3.2^3\)
\(72^{63}=\left(3^3.2^3\right)^{63}=3^{189}.2^{189}\)
Như vậy đề bài sai.
a) Xét từng vế ta có :
\(24^{54}.54^{24}.2^{10}=\left(2^3.3\right)^{54}.\left(2.3^2\right)^{24}.2^{10}\)
\(=2^{162}.3^{54}.2^{24}.3^{48}.2^{10}\)
\(=2^{172}.3^{102}\)
Xét vế tiếp theo ta có :
\(72^{63}=\left(2^3.3^2\right)^{63}=2^{189}.3^{126}\)
\(\Rightarrow72^{63}⋮24^{54}.2^{10}.54^{24}\)
\(\RightarrowĐPCM\)