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a)Đặt \(A=7^6+7^5-7^4\)
\(A=7^4\left(7^2+7-1\right)\)
\(A=7^4\cdot55⋮55\left(đpcm\right)\)
b)\(A=1+5+5^2+5^3+...+5^{50}\)
\(5A=5+5^2+5^3+5^4+...+5^{51}\)
\(5A-A=\left(5+5^2+5^3+5^4+...+5^{51}\right)-\left(1+5+5^2+5^3+...+5^{50}\right)\)
\(4A=5^{51}-1\)
\(A=\frac{5^{51}-1}{4}\)
a)
Ta có :
\(7^6+7^5-7^4=7^4\left(7^2+7-1\right)=7^4.55\)
=> Chia hết cho 5
b)
Ta có :
\(A=1+5+5^2+....+5^{50}\)
\(5A=5+5^2+....+5^{51}\)
=> 5A - A = \(\left(5+5^2+....+5^{51}\right)\)\(-\left(1+5+....+5^{50}\right)\)
\(\Rightarrow4A=5^{51}-1\)
\(\Rightarrow A=\frac{5^{51}-1}{4}\)
A = ( 7 . 1 + 7 . 49 ) + ( 7 . 7 + 7 . 343 )
A = 7 . 50 + 7 . 350
A = 7 . 400
Mà 400 \(⋮\)50 → A \(⋮\)50
A=7+72+73+74 = 7( 1+ 7+ 72+73) = 7( 1 + 7 + 49 + 343) = (7+400) ko chia hết cho 5
B=106-57 = 26. 56- 57 = 56( 26- 5) = 56 . 59
Do 56.59 chia hết cho 59 => B=..... chia hết cho 59
Bài 1 :
7^6+7^5-7^4=7^4.49+7^4.7-7^4.1
=7^4.(49+7-1)
=7^4.55
Vì 7^4.55 chia hết 5 Vậy 7^6+7^5-7^4 chia hết 5
C1:\(\left(2x+1\right)^3=125=5^3\Rightarrow2x+1=5\Rightarrow x=\left(5-1\right):2=2\)
C2:\(\left(3X-2\right)^4=81=3^4\Rightarrow3X-2=3\Rightarrow x=\left(3+2\right):3=\frac{5}{3}\)
C3:\(\left(4x-8\right)^3=64=4^3\Rightarrow4x-8=4\Rightarrow x=\left(4+8\right):4=3\)
2)\(A=7+7^2+7^3+7^4=7\left(1+7+7^2+7^3\right)=7.400=7.8.50\)
=> a chia hết cho 50
\(B=10^6-5^7=2^6.5^6-5^6.5=5^6\left(2^6-5\right)=5^6.59\Rightarrow A⋮59\)
a) x=4 b) x= \(\frac{5}{3}\)hoặc x=\(\frac{-1}{3}\) c) x=3
a) 76 + 75 - 74 = 74 ( 72 + 7 - 1) = 74 . 55\(⋮\)55
b) A = 1 + 5 + 52 + ... + 550
5A = 5 + 52 + 53 + ... + 551
5A - A = ( 5 + 52 + 53 + ... + 551) - ( 1 + 5 + 52 + ... + 550)
4A = 551 - 1
A = \(\frac{5^{51}-1}{4}\)
a) \(\frac{3^7.16^3}{12^5.27^2}=\frac{3^7.\left(4^2\right)^3}{3^5.4^5.\left(3^3\right)^2}\)
\(=\frac{3^7.4^6}{3^5.4^5.3^6}\)
\(=\frac{3^7.4^6}{3^{11}.4^5}\)
\(=\frac{4}{3^4}=\frac{4}{81}\)
b) \(\frac{2^3.\left(\frac{1}{2}\right)^3.3^7}{2.\left(\frac{1}{2}\right)^4.3^8}\)
\(=\frac{4}{\frac{1}{2}.3}=\frac{4}{\frac{3}{2}}=\frac{8}{3}\)
c)\(\frac{10^3+5.10^2+5^3}{-13}\)
\(=\frac{10^2.\left(10+5\right)+125}{-13}\)
\(=\frac{100.15+125}{-13}\)
\(=\frac{1500+125}{-13}=\frac{1625}{-13}=-125\)
A =7+73+75+...+71999
72A = 73 + 75 + 77 + ... + 72001
72A - A = ( 73 + 75 + 77 + ... + 72001 ) - ( 7+73+75+...+71999 )
48A = 72001 - 7
A = \(\frac{7^{2001}-7}{48}\)