C = 2009.2009

C = (2007+2).2009

C = 2007.2009 + 2.2009

D = 2007.2011

D = 2007.(2009+2)

D = 2007.2009 + 2.2007

Vì 2.2009 > 2.2007

=> 2007.2009 + 2.2009 > 2007.2009 + 2.2007

=> C > D

Bạn chỉ cần : 

tính nhanh 

A =1990² -1992.1998

B=2001² -2000.2002

C =2009²-2007.2011

D= (37.13-13):(24+37.12)

E = (16.17-5) :(16.16+11)

F= (45.16-17):(28+45.15)

ban phai chuyen ra thanh so tu nhien thi moi lam duoc.

E =16(16+1) -5 phan 16.16+11 =16.16+16 - 5phan 16.16 +11 = 16.16 +11 phan 16.16 =11

\(a,C=5+5^2+5^3+5^4+\cdot\cdot\cdot+5^{20}\)

\(=5\left(1+5+5^2+\cdot\cdot\cdot+5^{19}\right)\)

Ta thấy: \(5\left(1+5+5^2+\cdot\cdot\cdot+5^{19}\right)⋮5\)

nên \(C⋮5\)

\(b,C=5+5^2+5^3+5^4\cdot\cdot\cdot+5^{20}\)

\(=\left(5+5^2\right)+\left(5^3+5^4\right)+\cdot\cdot\cdot+\left(5^{19}+5^{20}\right)\)

\(=5\left(1+5\right)+5^3\left(1+5\right)+\cdot\cdot\cdot+5^{19}\left(1+5\right)\)

\(=5\cdot6+5^3\cdot6+\cdot\cdot\cdot+5^{19}\cdot6\)

\(=6\cdot\left(5+5^3+\cdot\cdot\cdot+5^{19}\right)\)

Ta thấy: \(6\cdot\left(5+5^3+\cdot\cdot\cdot+5^{19}\right)⋮6\)

nên \(C⋮6\)

\(c,C=5+5^2+5^3+5^4+\cdot\cdot\cdot+5^{20}\)

\(=\left(5+5^3\right)+\left(5^2+5^4\right)+\cdot\cdot\cdot+\left(5^{17}+5^{19}\right)+\left(5^{18}+5^{20}\right)\)

\(=5\left(1+5^2\right)+5^2\left(1+5^2\right)+\cdot\cdot\cdot+5^{17}\cdot\left(1+5^2\right)+5^{18}\left(1+5^2\right)\)

\(=5\cdot26+5^2\cdot26+\cdot\cdot\cdot+5^{17}\cdot26+5^{18}\cdot26\)

\(=26\cdot\left(5+5^2+\cdot\cdot\cdot+5^{17}+5^{18}\right)\)

Ta thấy: \(26\cdot\left(5+5^2+\cdot\cdot\cdot+5^{17}+5^{18}\right)⋮13\)

nên \(C⋮13\)

#\(Toru\)

a, ta có
C = 5 + 5^2 + 5^3 + 5^4 + ... + 5^20
=> C = 5 . ( 1 + 5 + 5^2 + 5^3 + ... + 5^19 )
=> C chia hết cho 5
b,
C = 5 + 5^2 + 5^3 + 5^4 + ... + 5^20
=> C = 5 . ( 1 + 5 ) + 5^3 . ( 1 + 5 ) + ... + 5^19 . ( 1 + 5 )
=> C = 5 . 6 + 5^3 . 6 + ... + 5^19 . 6
=> C = 6 . ( 5 + 5^3 + ... + 5^19 )
=> C chia hết cho 6
c,
C = 5 + 5^2 + 5^3 + ... + 5^20
=> C = (5 + 5^2 + 5^3 + 5^4 ) + ... + (5^17 + 5^18 + 5^19 + 5^20 )
=> C = 5 . ( 1 + 5 + 5^2 + 5^3 ) + ... + 5^17 . ( 1+ 5 + 5^2 +5^3)
=> C = 5 . 156 + 5^5 . 156 + ...+ 5^17 . 156
=> C = 5 . 12 . 13 + 5^5 . 12 . 13 + ... + 5^17 . 12 . 13
=> C = 13 . ( 5 . 12 + 5^5 . 12 + ... + 5^17 . 12 )
=> C chia hết cho 13

  

Ta có: a và b chia 5 dư 3

\(\Leftrightarrow\left\{{}\begin{matrix}a=5k+3\left(k\in N\right)\\b=5n+3\left(n\in N\right)\end{matrix}\right.\)

Ta có: c chia 5 dư 2

\(\Leftrightarrow c=5m+2\left(m\in N\right)\)

Ta có: a+c

\(=5k+3+5m+2\)

\(=5k+5m+5\)

\(=5\left(k+m+1\right)⋮5\)

Ta có: b+c

\(=5n+3+5m+2\)

\(=5n+5m+5\)

\(=5\left(n+m+1\right)⋮5\)

Ta có: a-b

\(=5k+3-\left(5n+3\right)\)

\(=5k+3-5n-3\)

\(=5k-5n\)

\(=5\left(k-n\right)⋮5\)

Super Man mà lại còn phải lên đây để hỏi bài à?

Super man hỏi bài? Nghịch lý

ok