Đặt M = \(1+9+9^2+......+9^{2010}\)

\(9M=9+9^2+9^3+......+9^{2011}\)

\(9M-M=8M=9^{2011}-1\)

Đặt K = \(1+9+9^2+......+9^{2009}\)

\(9K=9+9^2+9^3+.....+9^{2010}\)

\(9K-K=8K=9^{2010}-1\)

\(\Rightarrow A=\frac{9^{2011}-1}{9^{2010}-1}\)

Đặt H=\(1+5+5^2+....+5^{2010}\)

\(5H=5+5^2+......+5^{2011}\)

\(5H-H=4H=5^{2011}-1\)

ĐẶT G = \(1+5+5^2+.......+5^{2009}\)

\(5G-G=4G=5^{2010}-1\)

\(\Rightarrow B=\frac{5^{2011}-1}{5^{2010}-1}\)

Rồi bạn so sánh sẽ ra ngay

A = \(1+\frac{9^{2010}}{1+9+9^2+....+9^{2009}}\)= \(1+1:\frac{1+9+9^2+....+9^{2009}}{9^{2010}}\)= \(1+1:\left(\frac{1}{9^{2010}}+\frac{1}{9^{2009}}+\frac{1}{9^{2008}}+...+\frac{1}{9}\right)\)

B = \(1+\frac{5^{2010}}{1+5+5^2+....+5^{2009}}\)= \(1+1:\frac{1+5+5^2+...+5^{2009}}{5^{2010}}\)= \(1+1:\left(\frac{1}{5^{2010}}+\frac{1}{5^{2009}}+...+\frac{1}{5}\right)\)

Do \(\frac{1}{9^{2010}}

có đúng đề không vậy 

ai làm được cho 10 tick

a,Ta co:\(A=\frac{2005^{2005}+1}{2005^{2006}+1}<\frac{2005^{2005}+1+2004}{2005^{2006}+1+2004}=\frac{2005^{2005}+2005}{2005^{2006}+2005}\)

                 \(=\frac{2005\left(2005^{2004}+1\right)}{2005\left(2005^{2005}+1\right)}=\frac{2005^{2004}+1}{2005^{2005}+1}\) =B                                                                                        Vay A<B    

b,lam tuong tu nhu y a

\(a)\left(5^{2010}+5^{2012}+5^{2014}\right):\left(5^{2011}+5^{2009}+5^{2007}\right)\)

\(=\dfrac{5^{2007}\left(5^3+5^5+5^7\right)}{5^{2007}\left(5^4+5^2+1\right)}=\dfrac{5^3+5^5+5^7}{5^4+5^2+1}\)

\(=\dfrac{125+3125+78125}{625+25+1}=\dfrac{81375}{651}=125\)

\(b)-\dfrac{7}{45}+\dfrac{1}{4}+\dfrac{3}{5}+\dfrac{1}{12}+\dfrac{2}{3}+\dfrac{1}{39}+\dfrac{5}{9}\)

\(=\dfrac{-7.52+1.585+3.468+1.195+2.780+1.60-5.260}{2340}\)

\(=\dfrac{-364+585+1404+195+1560+60-1300}{2340}\)

\(=\dfrac{2140}{2340}=\dfrac{107}{117}\)

Ai lm đc câu nào thì giúp mk với , cảm ơn !!

\(A=\left|\dfrac{3}{5}-x\right|+\dfrac{1}{9}\ge\dfrac{1}{9}\\ A_{min}=\dfrac{1}{9}\Leftrightarrow x=\dfrac{3}{5}\\ B=\dfrac{2009}{2008}-\left|x-\dfrac{3}{5}\right|\le\dfrac{2009}{2008}\\ B_{max}=\dfrac{2009}{2008}\Leftrightarrow x=\dfrac{3}{5}\\ C=-2\left|\dfrac{1}{3}x+4\right|+1\dfrac{2}{3}\le1\dfrac{2}{3}\\ C_{max}=1\dfrac{2}{3}\Leftrightarrow\dfrac{1}{3}x=-4\Leftrightarrow x=-12\)

a: \(A=\left|\dfrac{3}{5}-x\right|+\dfrac{1}{9}\ge\dfrac{1}{9}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{3}{5}\)

A = 1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 + ... + 2005 + 2006 - 2007 - 2008 + 2009 + 2010 ( có 2010 số )

A = ( 1 + 2 - 3 - 4 ) + ( 5 + 6 - 7 - 8 ) + .... + ( 2005 + 2006 - 2007 - 2008 ) + ( 2009 + 2010 ) 

A = ( - 4 ) + ( - 4 ) + ... + ( - 4 ) + 4019 ( có 503 số )

A = ( - 4 ) . 502 + 4019

A = - 2008 + 4019

A = 2011.

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