Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
ta có: A=2^2009-2^2008-2^2007-......-2^2-2^1-2^0
=>2A=2^2010-2^2009-2^2007-........-2^2-2^1
=>2A-A=(2^2010-2^2009-2^2007-........-2^2-2^1)-(2^2009-2^2008-2^2007-......-2^2-2^1-2^0)
=>A = 2^2010-2^0
vậy A=2^2010-2^0
A=22009-22008-22007-....-22-21-20
A = 22009 - ( 22008 + 22007 + ... + 22 + 21 + 20 )
Đặt B = 22008 + 22007 + ... + 22 + 21 + 20
2B = 22009 + 22008 + 22007 + ... + 23 + 22 + 21
2B - B = ( 22009 + 22008 + 22007 + ... + 23 + 22 + 21 ) - ( 22008 + 22007 + ... + 22 + 21 + 20 )
B = 22009 - 20
Vậy A = 22009 - ( 22009 - 20 )
A = 22009- 22009 + 20
A = 20 = 1
1)Đặt A=1+2+22+23+.....+22008
=>2A=2+22+23+....+22009
=>2A-A=(2+22+23+...+22009)-(1+2+22+23+....+22008)
=-1+22009
Lời giải:
\(x=\frac{1}{2^{2009}}+\frac{2}{2^{2008}}+\frac{3}{2^{2007}}+....+\frac{2008}{2^2}+\frac{2009}{2}\)
\(2x = \frac{1}{2^{2008}}+\frac{2}{2^{2007}}+\frac{3}{2^{2006}}+...+\frac{2008}{2}+2009\)
\(\Rightarrow x=2x-x=2009-\frac{1}{2}-\frac{1}{2^2}-...-\frac{1}{2^{2008}}-\frac{1}{2^{2009}}\)
\(\Rightarrow 2009-x=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2008}}+\frac{1}{2^{2009}}\)
\(\Rightarrow 2(2009-x)=1+\frac{1}{2}+....+\frac{1}{2^{2007}}+\frac{1}{2^{2008}}\)
\(\Rightarrow 2(2009-x)-(2009-x)=1-\frac{1}{2^{2009}}\)
\(\Rightarrow 2009-x=1-\frac{1}{2^{2009}}\\ \Rightarrow x=2009-(1-\frac{1}{2^{2009}})=2008+\frac{1}{2^{2009}}\)
2A=2+2^2+....+2^51
A=2A-A=(2+2^2+...+2^51)-(1+2+2^2+...+2^50)=2^51-1
5B=5^2+5^3+.....+5^101
4B=5B-B=(5^2+5^3+....+5^101)-(5+5^2+...+5^100)=5^101-5
=> B=(5^101-5)/4
Tk mk nha
1)
X10=1X
=>x10=1
=>x10=110 ; x10=(-1)10
=>x=1;x=-1
2)
(2x-2007)2008-(2x-2007)2009=0
=>(2x-2007)2008.(1+2x-2007)=0
=>hoặc (2x-2007)2008=0 =>2x-2007=0 =>2x=2007 =>x=2007/2
hoặc 1+2x-2007=0 =>1+2x=2007 =>2x=2006 =>x=1003
Vậy x=2007/2 ; x=1003
3)
=>(x-1)2=1242
=>(x-1)2=? (Chịu)