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\(B=1+5+5^2+5^3+...+5^{2008}+5^{2009}\)
\(\Rightarrow5B=5\left(1+5+5^2+5^3+...+5^{2008}+2^{2009}\right)\)
\(5B=5+5^2+5^3+5^4+...+2^{2009}+2^{2010}\)\(\Rightarrow4B=5B-B=\left(5+5^2+5^3+5^4+...+5^{2009}+5^{2010}\right)-\left(1+5+5^2+5^3+...+5^{2008}+5^{2009}\right)\)
\(4B=5^{2010}-1\Leftrightarrow B=\dfrac{5^{2010}-1}{4}\) vậy \(B=\dfrac{5^{2010}-1}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
B=1+5+52+53+...+52008+52009 (1)
5B=5+52+53+54+...+52009+52010 (2)
Lấy (2) - (1) theo từng vế
5B - B = (5+52+53+54+....+52009+52010) - (1+5+52+53+...+52008+52009 )
<=> 5+52+53+54+...+52009+52010- 1-5-52-53-54-...-52008-52009
<=> (5-5)+(52-52)+(53-53)+...+(52009-52009)+(52010-1)
<=> 0+0+0+0+...+0)+0+52010-1
4B=52010-1
<=>B=(52010-1) : 4
![](https://rs.olm.vn/images/avt/0.png?1311)
C=(1-2-3+4)+(5-6-7+8)+...+(2005-2006-2007+2008)+2009-2010-2011
=-1-2011
=-2012
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có B=\(\frac{2009^{2010}-2}{2009^{2011}-2}\)<1
=>\(\frac{2009^{2010}-2}{2009^{2011}-2}\)<\(\frac{2009^{2010}-2+3}{2009^{2011}-2+3}\)=\(\frac{2009^{2010}+1}{2009^{2011}+1}\)(1)
Mà \(\frac{2009^{2010}+1}{2009^{2011}+1}\)<1
=> \(\frac{2009^{2010}+1}{2009^{2011}+1}\)<\(\frac{2009^{2010}+1+2008}{2009^{2011}+1+2008}\)=\(\frac{2009^{2010}+2009}{2009^{2011}+2009}\)=\(\frac{2009\cdot\left(2009^{2009}+1\right)}{2009\cdot\left(2009^{2010}+1\right)}\)=\(\frac{2009^{2009}+1}{2009^{2010}+1}\)=A(2)
Từ (1)và(2)=>B<\(\frac{2009^{2010}+1}{2009^{2011}+1}\)<A=>B<A hay A>B
![](https://rs.olm.vn/images/avt/0.png?1311)
\(P=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{19}}+\dfrac{1}{2^{20}}\\ 2P=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{18}}+\dfrac{1}{2^{19}}\\ 2P-P=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{18}}+\dfrac{1}{2^{19}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{19}}+\dfrac{1}{2^{20}}\right)\\ P=1-\dfrac{1}{2^{20}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Làm tạm một câu rồi đi chơi, lát làm cho.
4)
Áp dụng bất đẳng thức Cauchy-Schwarz :
\(VT\ge\frac{\left(1+1+1\right)^2}{a^2+b^2+c^2+2ab+2bc+2ca}=\frac{9}{\left(a+b+c\right)^2}\ge\frac{9}{1}=9\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=\frac{1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 2)
Ta có \(\frac{1}{a+1}+\frac{1}{b+1}\ge\frac{4}{3}\)
\(\Rightarrow\frac{b+1+a+1}{\left(a+1\right)\left(b+1\right)}\ge\frac{4}{3}\)
Ta có \(a+b=1\)
\(\Rightarrow\frac{3}{\left(a+1\right)\left(b+1\right)}\ge\frac{4}{3}\)
\(\Rightarrow\frac{3}{\left(a+1\right)b+a+1}\ge\frac{4}{3}\)
\(\Rightarrow\frac{3}{ab+b+a+1}\ge\frac{4}{3}\)
Ta có \(a+b=1\)
\(\Rightarrow\frac{3}{ab+2}\ge\frac{4}{3}\)
\(\Leftrightarrow9\ge4\left(ab+2\right)\)
\(\Rightarrow9\ge4ab+8\)
\(\Rightarrow1\ge4ab\)
Do \(a+b=1\Rightarrow\left(a+b\right)^2=1\)
\(\Rightarrow\left(a+b\right)^2\ge4ab\)
\(\Rightarrow a^2+2ab+b^2\ge4ab\)
\(\Rightarrow a^2-2ab+b^2\ge0\)
\(\Rightarrow\left(a-b\right)^2\ge0\) (đpcm )
Câu 3)
Ta có \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge9\)
Mà \(a+b+c=1\)
\(\Rightarrow\frac{a+b+c}{a}+\frac{a+b+c}{b}+\frac{a+b+c}{c}\ge9\)
\(\Rightarrow a+b+c\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)
Áp dụng bất đẳng thức Cô-si
\(\Rightarrow\left\{\begin{matrix}a+b+c\ge3\sqrt[3]{abc}\\\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\sqrt[3]{\frac{1}{abc}}\end{matrix}\right.\)
\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\sqrt[3]{abc}\sqrt[3]{\frac{1}{abc}}\)
\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9.\sqrt[3]{\frac{abc}{abc}}\)
\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\) (điều này luôn luôn đúng)
\(\Rightarrow\) ĐPCM
Đặt \(A=1+2+2^2+2^3+...+2^{2008}\)
\(2A=2\left(1+2+2^2+2^3+...+2^{2008}\right)\)
\(2A=2+2^2+2^3+...+2^{2009}\)
\(2A-A=\left(2+2^2+...+2^{2009}\right)-\left(1+2+...+2^{2008}\right)\)
\(A=2^{2009}-1=-\left(1-2^{2009}\right)\)
Khi đó \(B=\dfrac{A}{1-2^{2009}}=\dfrac{-\left(1-2^{2009}\right)}{1-2^{2009}}=-1\)
P/s: 1 .tìm trước khi hỏi
2. Nhớ để câu hỏi đúng box tái phạm thì đừng trách
2A ở đâu vậy?