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\(A=2^0+2^1+2^2+2^3+2^4+2^5+\dots+2^{100}\\=(2^1+2^2)+(2^3+2^4)+(2^5+2^6)+\dots+(2^{99}+2^{100})+2^0\\=2\cdot(1+2)+2^3\cdot(1+2)+2^5\cdot(1+2)+\dots+2^{99}\cdot(1+2)+1\\=2\cdot3+2^3\cdot3+2^5\cdot3+\dots+2^{99}\cdot3+1\\=3\cdot(2+2^3+2^5+\dots+2^{99})+1\)
Vì \(3\cdot(2+2^3+2^5+\dots+2^{99})\vdots3\)
\(\Rightarrow 3\cdot(2+2^3+2^5+\dots+2^{99})+1\) chia \(3\) dư 1
hay số dư của phép chia \(A\) cho \(3\) là \(1\).
A=2^0 + 2^1 + 2^2 + 2^3 + 2^4 + ....+2^100
A=1 + 2^1 + 2^2 + 2^3 + 2^4 + ....+2^100
A=1 + (2^1 + 2^2) + (2^3 + 2^4) + ....+(2^99 + 2^100)
A=1 + 2.(1+2) + 2^3.(1+2)+....+2^99.(1+2)
A=1 + 2 . 3 + 2^3 . 3 +....+2^99 . 3
A=1 +3 .(2+2^3+..+2^99)
=> A:3 dư 1
a)(x-2005).2006=0
=>x-2005=0
=>x=2005
b)480+45.4=(x+125):5+260
=>480+180=(x+125):5+260
=>660=(x+125):5+260
=>(x+125):5=400
=>x+125=2000
=>x=1875
c)2005.(x-2006)=2005
=>x-2006=1
=>x=2007
d){(x+50).50-50}:50=50
=>(x+50).50-50=2500
=>(x+50)*50=2550
=>x+50=51
=>x=1
a)x-2006=1=>x=2006
b)(x+50)*50-50=2500
(x+50)*50=2550
x+50=51
x=1
Tick Nha
[(x+50).50-50]:50=50
(x+50).50-50=50.50
(x+50).50-50=2500
(x+50).50=2500+50
(x+50).50=2550
x+50=2550:50
x+50=51
x=51-50
x=1
1) 50x50=2500
( x+50).50=2500+50=2550
x+50 =2550:50=51
x =51-50=1
nhầm
câu c :
=> 480 + 180 = (x + 125) : 5 + 260
=> 660 = (x + 125) : 5 + 260
=> (x + 125) : 5 = 400
=> x + 125 = 2000
=> x = 1875
câu e :
=> (x + 50) . 50 - 50 = 2500
=> (x + 50) . 50 = 2550
=> x + 50 = 51
=> x = 1
a/ = (195 - 195) : (1945 + 1014) = 0 : (1945 + 1014) = 0
b/ => x - 2005 = 0 => x = 0 + 2005 = 2005
c/ => 480 + 180 = (x + 125) : 5 + 260
=> 300 = (x + 125) : 5 + 260
=> (x + 125) : 5 = 40
=> x + 125 = 8
=> x = -117
d/ => x - 2006 = 1 => x = 2007
e/ => (x + 50) . 50 - 50 = 1
=> (x + 50) . 50 = 51
=> x + 50 = 51/50
=> x = -2449/50
\(x+50=100\)
\(x=100-50\)
\(x=50\)
x + 50 = 100
x = 100 - 50
x = 50