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a)Ta có
2^20-2^17 = 2^17 . 2^3 - 2^17 .1
= 2^17 .( 2^3 - 1)
= 2^17 . 7 chia hết cho 7
b)Ta có
10^6 + 5^7 =2^6 . 5^6 + 5^6 . 5
= 5^6. (2^6+ 5)
= 5^6 . 69
......
tới đây mink hết biết r bn tự giải tiếp đi nha!
\(M=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot\cdot\cdot\cdot\frac{9999}{10000}\)
\(=\frac{1.3}{2.2}\cdot\frac{2.4}{3.3}\cdot\frac{3.5}{4.4}\cdot\cdot\cdot\cdot\frac{99.101}{100.100}\)
\(=\frac{1}{2}\cdot\frac{101}{100}=\frac{101}{200}\)
Xét vế phải :
\(VP=\frac{99}{50}-\frac{97}{49}+...+\frac{7}{4}-\frac{5}{3}+\frac{3}{2}-1\)
\(=2.\left(\frac{99}{100}-\frac{97}{98}+...+\frac{7}{8}-\frac{5}{6}+\frac{3}{4}-\frac{1}{2}\right)\)
\(=2\left[\left(1-\frac{1}{100}\right)-\left(1-\frac{1}{98}\right)+...+\left(1-\frac{1}{4}\right)-\left(1-\frac{1}{2}\right)\right]\)
\(=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\right)\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{25}+\frac{1}{26}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+...+\frac{1}{25}\right)\)
\(=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{49}+\frac{1}{50}=VT\Rightarrow\left(đpcm\right)\)
1. \(A=\frac{1}{2}-\frac{2}{5}+\frac{1}{3}+\frac{5}{7}-\frac{-1}{6}+\frac{-4}{35}+\frac{1}{41}\)
\(=\frac{1}{2}-\frac{2}{5}+\frac{1}{3}+\frac{5}{7}+\frac{1}{6}-\frac{4}{35}+\frac{1}{41}\)
\(=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\right)-\left(\frac{2}{5}-\frac{5}{7}+\frac{4}{35}\right)+\frac{1}{41}\)
\(=\left(\frac{5}{6}+\frac{1}{6}\right)-\left(\frac{-11}{35}+\frac{4}{35}\right)+\frac{1}{41}\)\(=1-\frac{-7}{35}+\frac{1}{41}=1+\frac{1}{5}+\frac{1}{41}=\frac{251}{205}\)
2. a) \(1+4+4^2+4^3+......+4^{99}=\left(1+4\right)+\left(4^2+4^3\right)+.......+\left(4^{98}+4^{99}\right)\)
\(=\left(1+4\right)+4^2\left(1+4\right)+.........+4^{98}\left(1+4\right)\)
\(=5+4^2.5+........+4^{98}.5=5\left(1+4^2+.....+4^{98}\right)⋮5\)( đpcm )
b) \(3^{n+2}-2^{n+2}+3^n-2^n=\left(3^{n+2}+3^n\right)-\left(2^{n+2}+2^n\right)\)
\(=3^n\left(3^2+1\right)-2^n\left(2^2+1\right)=3^n\left(9+1\right)-2^n\left(4+1\right)\)
\(=3^n.10-2^n.5=3^n.10-2^{n-1+1}.5=3^n.10-2^{n-1}.2.5\)
\(=3^n.10-2^{n-1}.10=10\left(3^n-2^{n-1}\right)⋮10\)( đpcm )
1. A = 75(42004 + 42003 +...+ 42 + 4 + 1) + 25
A = 25 . [3 . (42004 + 42003 +...+ 42 + 4 + 1) + 1]
A = 25 . (3 . 42004 + 3 . 42003 +...+ 3 . 42 + 3 . 4 + 3 + 1)
A = 25 . (3 . 42004 + 3 . 42003 +...+ 3 . 42 + 3 . 4 + 4)
A = 25 . 4 . (3 . 42003 + 3 . 42002 +...+ 3 . 4 + 3 + 1)
A =100 . (3 . 42003 + 3 . 42002 +...+ 3 . 4 + 3 + 1) \(⋮\) 100
Ta có:
57+58+59
=57(1+5+52)
=57.31
Vì 31 chia hết cho 31=)57.31 chia hết cho 31
Vậy 57+58+59 chia hết cho 31
Học tốt nhé
c)\(^{5^7+5^8+5^9}\)
= \(5^7\left(1+5+5^2\right)\)
= \(5^7.31\)
\(5^7.31⋮31\)
\(\Rightarrow\)\(5^7+5^8+5^9\)\(⋮\)\(31\)
