2)Ta có: \(2^{332}< 2^{333}=\left(2^3\right)^{111}=8^{111}\)

              \(3^{223}>3^{222}=\left(3^2\right)^{111}=9^{111}\)

Vì \(8^{111}< 9^{111}\) mà \(2^{332}< 8^{111},3^{223}>9^{111}\) nên suy ra \(2^{332}< 3^{223}\)

Vậy \(2^{332}< 3^{223}\)

1) \(A=\dfrac{10^{2013}+1}{10^{2014}+1}\Rightarrow10A=\dfrac{10^{2014}+10}{10^{2014}+1}=\dfrac{10^{2014}+1}{10^{2014}+1}+\dfrac{9}{10^{2014}+1}=1+\dfrac{9}{10^{2014}+1}\)

\(B=\dfrac{10^{2014}+1}{10^{2015}+1}\Rightarrow10B=\dfrac{10^{2015}+10}{10^{2015}+1}=\dfrac{10^{2015}+1}{10^{2015}+1}+\dfrac{9}{10^{2015}+1}=1+\dfrac{9}{10^{2015}+1}\)Vì: \(10^{2014}+1< 10^{2015}+1\Rightarrow\dfrac{9}{10^{2014}+1}>\dfrac{9}{10^{2015}+1}\Rightarrow1+\dfrac{9}{10^{2014}+1}>1+\dfrac{9}{10^{2015}+1}\)

Nên suy ra \(10A>10B\Rightarrow A>B\)

Bạn thiếu đề rồi phải là trừ hay cộng j j chứ.

Xét:

`A+B=2+1/2+1/3+1/4+......+1/4026+1/3+1/5+1/7+......+1/4025`

`1/2+1/3+1/4+......+1/4026+1/3+1/5+1/7+......+1/4025>0`

`=>A+B>2`

Mà `1 2013/2014<2`

`=>A+B>1 2013/2014`

có \(\dfrac{2012+2013}{2013+2014}=\dfrac{2012}{2013+2014}+\dfrac{2013}{2013+2014}\)

mà\(\dfrac{2012}{2013+2014}< \dfrac{2012}{2013}\)

\(\dfrac{2013}{2013+2014}< \dfrac{2013}{2014}\)

\(\Rightarrow\dfrac{2012}{2013}+\dfrac{2013}{2014}>\dfrac{2012}{2013+2014}+\dfrac{2014}{2013+2014}\\ \Rightarrow\dfrac{2012}{2013}+\dfrac{2013}{2014}>\dfrac{2012+2013}{2013+2014}\\ \Rightarrow A>B\)

\(A=-\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)...\left(1-\dfrac{1}{2014^2}\right)\)

\(A=\dfrac{\left(1\cdot3\right)\left(2\cdot4\right)\left(3\cdot5\right)...\left(2012\cdot2014\right)\left(2013\cdot2015\right)}{\left(2\cdot2\right)\left(3\cdot3\right)\left(4\cdot4\right)...\left(2013\cdot2013\right)\left(2014\cdot2014\right)}\)

\(A=\dfrac{\left(1\cdot2\cdot3\cdot...\cdot2012\cdot2013\right)\left(3\cdot4\cdot5\cdot...\cdot2014\cdot2015\right)}{\left(2\cdot3\cdot4\cdot...\cdot2013\cdot2014\right)\left(2\cdot3\cdot4\cdot...\cdot2013\cdot2014\right)}\)

\(A=\dfrac{1\cdot2015}{2014\cdot2}=\dfrac{2015}{4028}\)

Vì \(\dfrac{2015}{4028}>-\dfrac{1}{2}\) nên A > B

Ai trả lời đi please

A= 1+(\(\dfrac{1}{2014}\)+1)+(\(\dfrac{2}{2013}\)+1)+...+(\(\dfrac{2013}{2}\)+1)

= \(\dfrac{2015}{2015}\)+(\(\dfrac{1}{2014}\)+1)+(\(\dfrac{2}{2013}\)+1)+...+(\(\dfrac{2013}{2}\)+1)

= 2015.(\(\dfrac{1}{2015}\)+\(\dfrac{1}{2014}\)+\(\dfrac{1}{2013}\)+...+\(\dfrac{1}{2}\))=2015.B

\(\Rightarrow\) \(\dfrac{A}{B}\)=2015

Ta có : 

A=\(\frac{9}{a^{2013}}+\frac{7}{a^{2014}}\)

  =\(\left(\frac{8}{a^{2013}}+\frac{1}{a^{2013}}\right)+\left(\frac{8}{a^{2014}}-\frac{1}{a^{2014}}\right)\)

=\(\left(\frac{8}{a^{2013}}+\frac{8}{a^{2014}}\right)+\left(\frac{1}{a^{2013}}-\frac{1}{a^{2014}}\right)\)

B=\(\frac{8}{a^{2014}}+\frac{8}{a^{2013}}\)

  =\(\frac{8}{a^{2013}}+\frac{8}{a^{2014}}\)

 Vì \(\frac{1}{a^{2013}}>\frac{1}{a^{2014}}\)nên\(\frac{1}{a^{2013}}-\frac{1}{a^{2014}}>0\)

=> \(\left(\frac{8}{a^{2013}}+\frac{8}{a^{2014}}\right)+\left(\frac{1}{a^{2013}}-\frac{1}{a^{2014}}\right)>\frac{8}{a^{2013}}+\frac{8}{a^{2014}}\)

Vậy \(A>B\)

Chúc em học tốt

#Thiên_Hy

\(A=\frac{9}{a^{2013}}+\frac{7}{a^{2014}}=\frac{8}{a^{2013}}+\frac{1}{a^{2013}}+\frac{7}{a^{2014}}\)

\(B=\frac{8}{a^{2014}}+\frac{8}{a^{2013}}=\frac{7}{a^{2014}}+\frac{1}{a^{2014}}+\frac{8}{a^{2013}}\)

Ta thấy :

\(\frac{8}{a^{2013}}=\frac{8}{a^{2013}}\)

\(\frac{7}{a^{2014}}=\frac{7}{a^{2014}}\)

\(\frac{1}{a^{2013}}>\frac{1}{a^{2014}}\left(a^{2013}< a^{2014}\right)\)

\(\Rightarrow A>B\)

\(A=\dfrac{2014^{2013}+1}{2014^{2014}+1}\Leftrightarrow2014A=\dfrac{2014^{2014}+2014}{2014^{2014}+1}=\dfrac{2014^{2014}+1+2013}{2014^{2014}+1}=1+\dfrac{2013}{2014^{2014}+1}\)

\(B=\dfrac{2014^{2012}+1}{2014^{2013}+1}\Leftrightarrow2014B=\dfrac{2014^{2013}+2014}{2014^{2013}+1}=\dfrac{2014^{2013}+1+2013}{2014^{2013}+1}=1+\dfrac{2013}{2014^{2013}+1}\)

Dễ thấy: \(1+\dfrac{2013}{2014^{2014}+1}< 1+\dfrac{2013}{2014^{2013}+1}\) nên \(2014A< 2014B\) hay \(A< B\)