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\(A=2^1+2^2+2^3+...+2^{59}+2^{60}.\)
\(A=\left(2^1+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{58}+2^{59}+2^{60}\right).\)
\(A=2^1\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{58}\left(1+2+2^2\right).\)
\(A=2^1\left(1+2+4\right)+2^4\left(1+2+4\right)+...+2^{58}\left(1+2+4\right).\)
\(A=2^1.7+2^4.7+...+2^{58}.7.\)
\(A=\left(2^1+2^4+...+2^{58}\right).7⋮7\left(đpcm\right).\)
A = 21 + 22 + 23 + ..... + 259 + 260
A = ( 21 + 22 + 23 ) + ... + ( 258 + 259 + 260 )
A = 21 . ( 1 + 2 + 22 ) + ... + 258 . ( 1 + 2 + 22 )
A = 21 . 7 + ... + 258 . 7 \(⋮\)7
Vậy A \(⋮\) 7
\(\dfrac{2}{1.2}+\dfrac{2}{2.3}+\dfrac{2}{3.4}+...............+\dfrac{2}{2008.2009}\)
\(=2\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+................+\dfrac{1}{2008.2009}\right)\)
\(=2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+.................+\dfrac{1}{2008}-\dfrac{1}{2009}\right)\)
\(=2\left(1-\dfrac{1}{2009}\right)\)
\(=2.\dfrac{2008}{2009}=\dfrac{4016}{2009}\)
A = ( 1 + 6 + 6^2 ) + ( 6^3 + 6^4 + 6^5 ) + ... + ( 6^57 + 6^58 + 6^59 )
= 1( 1 + 6 + 6^2 ) + 6^3( 1 + 6 + 6^2 ) + ... + 6^57( 1 + 6 + 6^2 )
= 1.43 + 6^3.43 + ... + 6^57.43
= 43( 1 + 6^3 + ... + 6^57 )
=> A chia hết cho 43
A = ( 1 + 6 ) + ( 6^2 + 6^3 ) + ... + ( 6^58 + 6^59 )
= 1( 1 + 6 ) + 6^2( 1 + 6 ) + ... + 6^58( 1 + 6 )
= 1.7 + 6^2.7 + ... + 6^58.7
= 7( 1 + 6^2 + ... + 6^58 )
=> A chia hết cho 7
Bg
a)\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.....\frac{99^2}{99.100}.\frac{100^2}{100.101}\)
\(=\frac{1^2.2^2.3^2.....99^2.100^2}{1.2.2.3.3.4.....99.100.100.101}\)
\(=\frac{1^2}{101}\)
\(=\frac{1}{101}\)
Ghi chú: \(=\frac{1^2.2^2.3^2.....99^2.100^2}{1.2.2.3.3.4.....99.100.100.101}\)--> 22 chịt tiêu 2.2 (trên và dưới) làm thế này mãi đến khi còn \(\frac{1^2}{101}\).
b) \(\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.....\frac{59^2}{58.60}\)
=\(\frac{2^2.3^2.4^2.....59^2}{1.3.2.4.3.5.....58.60}\)
= \(\frac{2}{1}.\frac{59}{60}\)
= \(\frac{59}{30}\)
Ghi chú: \(\frac{2^2.3^2.4^2.....59^2}{1.3.2.4.3.5.....58.60}\)--> chịt tiêu liên tục, còn \(\frac{2}{1}.\frac{59}{60}\).
bai nay minh la the nay cac ban doc neu cach lam dung thi tk giup neu sai thi nhan tin cho minh
Giai
Nhan ca hai ve voi 2 ta co :
\(2A=2.2^1.2^2.2^3.....................2^{59}.2^{60}\)
\(-\)
\(A=2.^12^2.2^3..................2^{59}.2^{60}\)
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\(2A=2\)
\(\Rightarrow A=1\)
Vi \(1⋮7\)
\(\Rightarrow A=2^1.2^2.2^3...................2^{59}.2^{60}⋮7\)
sai zùi