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![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+...+\frac{2014}{2^{2014}}\)
\(\Rightarrow2A=1+\frac{2}{2}+\frac{3}{2^2}+...+\frac{2014}{2^{2013}}\)
\(\Rightarrow2A-A=\left(1+\frac{2}{2}+\frac{3}{2^2}+...+\frac{2014}{2^{2013}}\right)-\left(\frac{1}{2}+\frac{2}{2^2}+...+\frac{2014}{2^{2014}}\right)\)
\(\Rightarrow A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2013}}-\frac{2014}{2^{2014}}\)
Đặt \(B=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2013}}\)
\(\Rightarrow2B=2+1+...+\frac{1}{2^{2012}}\)
\(\Rightarrow2B-B=\left(2+1+...+\frac{1}{2^{2012}}\right)-\left(1+\frac{1}{2}+...+\frac{1}{2^{2013}}\right)\)
\(\Rightarrow B=2-\frac{1}{2^{2013}}< 2\)
\(\Rightarrow B< 2\)
\(\Rightarrow A< 2-\frac{2014}{2^{2014}}< 2\)
\(\Rightarrow A< 2\left(đpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2013}-\frac{1}{2014}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2A=2^{2015}-2^{2014}-...-2^2-2\)
\(2A-A=2^{2015}+1>2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A=-1/2*-2/3*-3/4*..*-2013/2014
A=-1*-2*-3*...*-2013/2*3*4*...*2014
A=-1/2014
ta có(-1)^2015=-1
B=-1/2015>-1/2014=A
nên A<B
![](https://rs.olm.vn/images/avt/0.png?1311)
A<1 vì 14^15+3<14^16+3 mà B>1 vì 2016^2014+1>2016^2013+1
nên A<B
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{24\cdot47-23}{24+47\cdot23}.\frac{3+\frac{3}{7}-\frac{3}{11}+\frac{3}{1001}-\frac{3}{13}}{\frac{9}{1001}-\frac{9}{13}+\frac{9}{7}-\frac{9}{11}+9}\)
\(=\frac{24\cdot\left(24+23\right)-23}{24+\left(24+23\right)\cdot23}\cdot\frac{3\left(1+\frac{1}{7}-\frac{1}{11}+\frac{1}{1001}-\frac{1}{13}\right)}{9\left(\frac{1}{1001}-\frac{1}{13}+\frac{1}{7}-\frac{1}{11}+1\right)}\)
\(=\frac{24^2+24\cdot23-23}{24+24\cdot23+23^2}\cdot\frac{3}{9}\) \(=\frac{24^2+23\cdot\left(24-1\right)}{\left(23+1\right)\cdot24\cdot23^2}\cdot\frac{1}{3}=1\cdot\frac{1}{3}=\frac{1}{3}\)
A=\(\frac{2014}{2014^a}+\frac{2014}{2014^b}\)=B=\(\frac{2013}{2015^a}\)+\(\frac{2015}{2013^b}\)
Ta có: 2014/\(2014^a\)+2014/2014^b= 2013/2014^a + 1/2014^a +2015/2014^a - 1/2014^a
=(2013/2014^a + 2015/2014^b) + ( 1/2014^a + 1/2014^b)
= B + (1/2014^a + 1/2014^b)
*Nếu a=b thì A=B
*Nếu a>b thì (1/2014^a + 1/2014^b) >0
\(\Rightarrow\) A< B
*Nếu a<b thì (1/2014^a + 1/2014^b)>0
\(\Rightarrow\) A>B