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A=4+(22+23+24+...+220)
A-4=22+23+24+...+220
2(A-4)=23+24+25+...+221
A-4=2(A-4)-(A-4)=(23+24+25+...+221)-(22+23+24+...+220)
A-4=(23-23)+(24-24)+(25-25)+...+(220-220)+(221-22)
A-4=221-4
A =221-4+4
A =221
Bạn làm tiếp nha .
1. \(A=2^{2016}-1\)
\(2\equiv-1\left(mod3\right)\\ \Rightarrow2^{2016}\equiv1\left(mod3\right)\\ \Rightarrow2^{2016}-1\equiv0\left(mod3\right)\\ \Rightarrow A⋮3\)
\(2^{2016}=\left(2^4\right)^{504}=16^{504}\)
16 chia 5 dư 1 nên 16^504 chia 5 dư 1
=> 16^504-1 chia hết cho 5
hay A chia hết cho 5
\(2^{2016}-1=\left(2^3\right)^{672}-1=8^{672}-1⋮7\)
lý luận TT trg hợp A chia hết cho 5
(3;5;7)=1 = > A chia hết cho 105
2;3;4 TT ạ !!
Ta có :
\(A=1+5+5^2+...+5^{32}\)
\(A=\left(1+5+5^2\right)+\left(5^3+5^4+5^5\right)+...+\left(5^{30}+5^{31}+5^{32}\right)\)
\(A=31+5^3\left(1+5+5^2\right)+...+5^{30}\left(1+5+5^2\right)\)
\(A=31+31.5^3+...+31.5^{30}\)
\(A=31\left(1+5^3+...+5^{30}\right)\) chia hết cho 31
Vậy \(A\) chia hết cho 31
\(a)\) Ta có :
\(\frac{a}{b}< \frac{a+c}{b+c}\)
\(\Leftrightarrow\)\(a\left(b+c\right)< b\left(a+c\right)\)
\(\Leftrightarrow\)\(ab+ac< ab+bc\)
\(\Leftrightarrow\)\(ac< bc\)
\(\Leftrightarrow\)\(a< b\)
Mà \(a< b\) \(\Rightarrow\) \(\frac{a}{b}< 1\)
Vậy ...
\(A\frac{27^4.8^{17}}{9^6.32^3}=\frac{\left(3^3\right)^4.\left(2^3\right)^{17}}{\left(3^2\right)^6.\left(2^5\right)^3}=\frac{3^{12}.2^{51}}{3^{12}.2^{15}}=\frac{3^{12}.2^{15}.2^{36}}{3^{12}.2^{15}}=2^{36}\)
\(B=\frac{72^3.54^3:8^3}{108^5:4^5}=\frac{\left(72.54:8\right)^3}{\left(108:4\right)^5}=\frac{486^3}{27^5}=\frac{\left(3^5.2\right)^3}{\left(3^3\right)^5}=\frac{3^{15}.2^3}{3^{15}}=2^3=8\)
Bài 2
A = 2 +22 + 23 + 24 + ....+ 2100
A = ( 2+22 ) + (23 + 24 ) + ....+ (299 + 2100 )
A = 2(1+2 ) + 23 (1+2 ) + ...+ 299(1+2)
A = 2.3 + 23.3 + ....+ 299 .3
A = 3(2+23 + ...+ 299 )
=> A \(⋮\) 3 ( đpcm )
Bài 3
a, 2.3x = 312 .34 + 20 .274
2.3x = 312 . 34 + 20 . (33 ) 4
2.3x = 312 .34 + 20 .312
2.3x = 312(34+20 )
2.3x = 312 . 54
2.3x = 312 . 27 .2
2.3x = 312 . 33 .2
2.3x = 315 .2
=> x=15
b , (2x +1 ) 2 + 3.(22 + 1 ) = 22 .10
(2x +1 ) 2 + 3.(4+1 ) = 4.10
(2x +1 ) 2 + 3.5 = 40
(2x +1 ) 2 + 15 = 40
(2x +1 ) 2 = 40-15
(2x +1 ) 2 = 25
(2x +1 ) 2 = 52
=> 2x + 1 = 5
2x = 5-1
2x = 4
2x = 22
=> x=2
\(A=5+5^2+5^3+5^4+........+5^{2010}\)
A = ( 1 + 5 + 52 ) + ............ + ( 52008 + 52009 + 52010 )
A = 31 + ......... + 31( 1 + 5 + 52 )
Mà 31\(⋮\)31 => A \(⋮\)31 ( đpcm )
a, 4 + \(4^2\) + \(4^3\) + ... + \(4^{60}\) chia hết cho 5
= ( 4 + \(4^2\) ) + ( \(4^3\) + \(4^4\) ) +... + ( \(4^{59}\) + \(4^{60}\))
= ( 4 + \(4^2\) ) + \(4^3\) . ( 4 + \(4^2\) ) +... + \(4^{59}\). ( 4 + \(4^2\) )
= 20 + \(4^3\) . 20 + ... + \(4^{59}\) . 20
= 20 . ( 1 + \(4^3\) + ... + \(4^{59}\) ) chia hết cho 5
4 + \(4^2\) + \(4^3\) + ... + \(4^{60}\) chia hết cho 21
= ( 4 + \(4^2\) + \(4^3\) ) + ( \(4^4\) + \(4^5\) + \(4^6\) ) + ... + ( \(4^{58}\)+ \(4^{59}\) + \(4^{60}\) )
= ( 4 + \(4^2\) + \(4^3\) ) + \(4^4\) . ( 4 + \(4^2\) + \(4^3\) ) + ... + \(4^{58}\) . ( 4 + \(4^2\) + \(4^3\) )
= 84 + \(4^4\) . 84 + .... + \(4^{58}\) . 84
= 84 . ( 1 + \(4^4\) + ... + \(4^{58}\) ) chia hết cho 21
b, 5 + \(5^2\) + \(5^3\) + ... + \(5^{10}\) chia hết cho 6
= ( 5 + \(5^2\) ) + ( \(5^3\) + \(5^4\) ) + ... + ( \(5^9\) + \(5^{10}\) )
= ( 5 + \(5^2\) ) + \(5^3\) . ( 5 + \(5^2\) ) + ... + \(5^9\) . ( 5 + \(5^2\) )
= 30 + \(5^3\) . 30 + ... + \(5^9\) . 30
= 30 . ( 1 + \(5^3\) + ... + \(5^9\) ) chia hết cho 6
b) \(A=1+5+5^1+5^2+5^3+...+5^{71}\)
\(\Rightarrow A=\left(1+5^1+5^2\right)+5^3\left(1+5^1+5^2\right)+...+5^{69}\left(1+5^1+5^2\right)\)
\(\Rightarrow A=31+5^3.31+...+5^{69}.31\)
\(\Rightarrow A=31\left(1+5^3+...+5^{69}\right)⋮31\left(dpcm\right)\)
a) \(A=1+5^1+5^2+5^3+...+5^{71}\)
\(\Rightarrow A=\dfrac{5^{71+1}-1}{5-1}=\dfrac{5^{72}-1}{4}\)
\(4A+x=5^{72}\)
\(\Rightarrow4.\dfrac{5^{72}-1}{4}+x=5^{72}\)
\(\Rightarrow5^{72}-1+x=5^{72}\)
\(\Rightarrow x=1\)