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26 tháng 10 2018
\(2^{225}=\left(2^3\right)^{75}=8^{75}< 9^{75}=\left(3^2\right)^{75}=3^{150}\)
\(2^{2009}+2^{2008}+.......+2+1=b\)
\(\Rightarrow2b=2^{2010}+2^{2009}+.........+2^2+2\)
\(\Rightarrow2b-b=2^{2010}-1\Rightarrow b=2^{2010}-1\)
\(\Rightarrow A=2^{2010}-b=2^{2010}-\left(2^{2010}-1\right)=1\)
22 tháng 10 2018
a, Đặt \(A=2^{2010}+2^{2009}+2^{2008}+...+2^1+2^0\)
\(\Rightarrow2A=2^{2011}+2^{2010}+2^{2009}+...+2^2+2^1\)
\(\Rightarrow2A-A=2^{2011}-2^0\)
\(\Rightarrow A=2^{2011}-1\)
b,\(7^{x+2}+2.7^{x-1}=345\)
\(7^{x-1}.\left(7^3+2\right)=345\)
\(\Rightarrow7^{x-1}.345=345\)
\(\Rightarrow7^{x-1}=345:345=1\)
\(\Rightarrow7^{x-1}=7^0\)
\(\Rightarrow x-1=0\)
\(\Rightarrow x=1\)
Vậy \(x=1\)
TK
1
LM
17 tháng 3 2018
Ta có: \(H=2^{2010}-2^{2009}-2^{2008}-...-2-1\)
\(=2^{2010}-\left(2^{2009}+2^{2008}+...+2+1\right)\)
Đặt \(A=2^{2009}+2^{2008}+...+2+1\)
\(\Rightarrow2A=2^{20010}+2^{2009}+...+2^2+2\)
\(\Rightarrow2A-A=\left(2^{20010}+2^{2009}+...+2^2+2\right)-\left(2^{2009}+2^{2008}+...+2+1\right)\)\(\Rightarrow A=\left(2^{2010}-1\right)+\left(2^{2009}-2^{2009}\right)+\left(2^{2008}-2^{2008}\right)+...+\left(2-2\right)\)\(\Rightarrow A=2001-1\)
\(\Rightarrow H=2^{2010}-\left(2^{2010}-1\right)\)
\(\Rightarrow H=2^{2010}-2^{2010}+1=1\)
Thay \(H=1\) vào biểu thức \(2010^H\)
\(\Rightarrow2010^H=2010^1=1\)
Vậy \(2010^H=1\)
NM
2
AH
Akai Haruma
Giáo viên
28 tháng 2 2020
Lời giải:
$\frac{x-1}{2011}+\frac{x-2}{2010}-\frac{x-3}{2009}=\frac{x-4}{2008}$
$\Leftrightarrow \frac{x-1}{2011}+\frac{x-2}{2010}=\frac{x-3}{2009}+\frac{x-4}{2008}$
$\Leftrightarrow \frac{x-1}{2011}-1+\frac{x-2}{2010}-1=\frac{x-3}{2009}-1+\frac{x-4}{2008}-1$
$\Leftrightarrow \frac{x-2012}{2011}+\frac{x-2012}{2010}=\frac{x-2012}{2009}+\frac{x-2012}{2008}$
$\Leftrightarrow (x-2012)\left(\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\right)=0$
Dễ thấy $\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}< 0$
Do đó $x-2012=0\Rightarrow x=2012$
VM
28 tháng 2 2020
\(\frac{x-1}{2011}+\frac{x-2}{2010}-\frac{x-3}{2009}=\frac{x-4}{2008}.\)
\(\Rightarrow\frac{x-1}{2011}+\frac{x-2}{2010}=\frac{x-4}{2008}+\frac{x-3}{2009}\)
\(\Rightarrow\left(\frac{x-1}{2011}-1\right)+\left(\frac{x-2}{2010}-1\right)=\left(\frac{x-4}{2008}-1\right)+\left(\frac{x-3}{2009}-1\right)\)
\(\Rightarrow\left(\frac{x-1-2011}{2011}\right)+\left(\frac{x-2-2010}{2010}\right)=\left(\frac{x-4-2008}{2008}\right)+\left(\frac{x-3-2009}{2009}\right)\)
\(\Rightarrow\frac{x-2012}{2011}+\frac{x-2012}{2010}=\frac{x-2012}{2008}+\frac{x-2012}{2009}\)
\(\Rightarrow\frac{x-2012}{2011}+\frac{x-2012}{2010}-\frac{x-2012}{2008}-\frac{x-2012}{2009}=0\)
\(\Rightarrow\left(x-2012\right).\left(\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2008}-\frac{1}{2009}\right)=0\)
Vì \(\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2008}-\frac{1}{2009}\ne0.\)
\(\Rightarrow x-2012=0\)
\(\Rightarrow x=0+2012\)
\(\Rightarrow x=2012\)
Vậy \(x=2012.\)
Chúc bạn học tốt!
PN
1
31 tháng 5 2017
Đặt \(A=2^{2009}+2^{2008}+...+2+2^0\)
\(=1+2+...+2^{2008}+2^{2009}\)
\(\Rightarrow2A=2+2^2+...+2^{2010}\)
\(\Rightarrow2A-A=\left(2+2^2+...+2^{2010}\right)-\left(1+2+...+2^{2009}\right)\)
\(\Rightarrow A=2^{2010}-1\)
\(\Rightarrow M=2^{2010}-\left(2^{2010}-1\right)\)
\(=2^{2010}-2^{2010}+1=1\)
Vậy M = 1
Đặt A = 22009 + 22008 + ... + 21 + 20. Khi đó, M = 22010 - A
Ta có 2A = 22010 + 22009 + ... + 22 + 21.
Suy ra 2A - A = 22010 - 20 = 22010 - 1.
Do đó M = 22010 - A = 22010 - (22010 - 1) = 22010 - 22010 + 1 = = 1.
M=2^2010-(2^2009+2^2008+2^2007+...+2^1+2^0)
M=22010-22009-22008-22007-...-21-20
=>2M=22011-22010-22009-22008-...-22-21
=>2M-M=22011-22010-22009-22008-...-22-21-(22010-22009-22008-22007-...-21-20)
=>M=22011-22010-22009-22008-...-22-21-22010+22009+22008+22007+...+21+20
=22011-22010-22010+20
=22011-2.22010+1
=22011-22011+1
=1
vậy M=1