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\(81^7-27^9-9^{13}\)
\(=\left(3^4\right)^7-\left(3^3\right)^9-\left(3^2\right)^{13}\)
\(=3^{28}-3^{27}-3^{26}\)
\(=3^{26}\left(3^2-3-1\right)\)
\(=3^{26}.5\)
\(=3^{22}.3^4.5=3^{22}.405⋮405\)
\(12^{2n+1}+11^{n+2}\)
\(=144^n.12+11^n.121\)
\(=144^n.12-11^n.12+11^n.133\)
\(=\left(144^n-11^n\right).12+11^n.133\)
Ta có: \(a^n-b^n⋮a-b\Rightarrow144^n-11^n⋮133\)
Vậy \(12^{2n+1}+11^{n+2}⋮133\)
Lời giải:
1)
Ta có : \(A=81^7-27^9-9^{13}=(3^4)^7-(3^3)^9-(3^2)^{13}\)
\(\Leftrightarrow A=3^{28}-3^{27}-3^{26}=3^{26}(3^2-3-1)\)
\(\Leftrightarrow A=5.3^{26}=405.3^{22}\)
Do đó \(A\vdots 405\) (đpcm)
2)
Ta thấy : \(12^{2}\equiv 11\pmod {133}\)
\(\Rightarrow 12^{2n+1}\equiv 11^{n}.12\pmod {133}\)
\(\Rightarrow 12^{2n+1}+11^{n+2}\equiv 11^n.12+11^{n+2}\pmod {133}\)
\(\Leftrightarrow 12^{2n+1}+11^{n+2}\equiv 11^n(12+11^2)\equiv 11^n.133\equiv 0\pmod {133}\)
Do đó: \(12^{2n+1}+11^{n+2}\vdots 133\) (đpcm)
3)
Ta thấy \(A=5x+2y;B=9x+7y\Rightarrow 3A+4B=51x+34y\)
Vì \(51\vdots 17;34\vdots 17\Rightarrow 3A+4B\vdots 17\)
Nếu \(A\vdots 17\Rightarrow 4B\vdots 17\). Mà $(4,17)$ nguyên tố cùng nhau nên \(B\vdots 17\)
Do đó ta có đpcm.
a: Ta có: \(\left(8\cdot5^7+5^6-5^5\right):5^5\)
\(=8\cdot5^2+5-1\)
\(=200+4=204\)
b: Ta có: \(\left(9^{30}-27^{19}\right):3^{57}+\left(125^9-25^{12}\right):5^{24}\)
\(=3^{60}:3^{57}-3^{57}:3^{57}+5^{27}:5^{24}-5^{24}:5^{24}\)
\(=27-1+125-1\)
=150
a. (8,57 - 55 + 56) : 55
= (8,57 : 55) - (55 : 55) + (56 : 55)
= 1,72 - 1 + 5
= 2,89 - 1 + 5
= 6,89
b. (930 - 2719) : 357 + (1259 - 2512) : 524
= (930 : 357) - (2719 : 357) + (1259 : 524) - (2512 : 524)
= 33 - 1 + 125 - 1
= 27 - 1 + 125 - 1
= 150
c. (1012 + 511 . 29 - 513 - 28) : 4 . 55 . 106
= (1012 + 2,5 , 1010 - 513 - 28) : 1,25 . 1010
= (1012 : 1,25 . 1010) + (2,5 . 1010 : 1,25 . 1010) - (513 : 1,25 . 1010) - (28 : 1,25 . 1010)
= 80 + 2 - \(\dfrac{25}{256}\) - \(\dfrac{1}{48828125}\)
= 81,90234373 \(\approx\) 82
\(11^{n+2}+12^{2n+1}=121.11^n+12.144^n\)
= ( 133 - 12 ) . \(11^n\)+ 12.\(144^n\)= 133 .11\(^n\)+ ( 144 \(^n-11^n\)) .12
Ta có : 133 . \(11^n\)chia hết cho 133 ; 144\(^n-11^n\)chia hết cho ( 144 - 11 )
=> 144\(^n-11^n\)chia hết cho 133
Ta có: 12
2n+1 + 11n+2
= 122n.12 + 11n.112
= 144n.12 + 11n.121
= 144n.12 - 11n.12 + 11n.121 + 11n.12
= 12.(144n - 11n) + 11n.(121 + 12)
= 12.(144n - 11n) + 11n.133
Do 144n - 11n luôn chia hết cho 144 - 11 = 133 => 12.(144n - 11n) chia hết cho 133; 11n.133 chia hết cho 133
=> 122n+1 + 11n+2 chia hết cho 133 ( đpcm)
\(C=x^{14}-10x^{13}+10x^{12}-10x^{11}+...+10x^2-10x+10\)
\(=x^{14}-\left(x+1\right)x^{13}+\left(x+1\right)x^{12}-\left(x+1\right)x^{11}+..+\left(x+1\right)x^2-\left(x+1\right)x+x+1\)
\(=x^{14}-x^{14}-x^{13}+x^{13}+x^{12}-x^{12}-x^{11}+...+x^3+x^2-x^2-x+x+1\)
\(=1\)
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 )