mk nghĩ là a>b

khó quá

A<B đó 

đảm bao 100% luôn

bạn ghi cách giải ra giúp mình với

A<B đó đúng 100%

\(A=2^0+2^1+2^2+2^3+...+2^{2010}\)

\(A=1+2+2^2+2^3+...+2^{2010}\)

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

\(2A-A=\left[2+2^2+2^3+...+2^{2011}\right]-\left[1+2+2^2+2^3+...+2^{2010}\right]\)

\(A=2^{2011}-1\)

Mà \(B=2^{2011}-1\)

=> A = B

Ta có: A=\(2^0+2^1+2^2+2^3+...+2^{2010}\)

          2A=\(2^1+2^2+2^3+2^4+...+2^{2011}\)

     2A-A hay A=\(2^{2011}-2^0\)

                       =\(2^{2011}-1\)

Vì \(2^{2011}-1=2^{2011}-1\)

\(\Rightarrow\)A=B

Hok tốt nha!!!

`A``=``2^0``+`2^1``+``2^2``+`2^3``+`...`+``2^(2010)`

`2A=2^1+2^2+2^3+2^4+...+2^(2011)`

`2A-A=(2^1+2^2+2^3+2^4+...+2^(2011))-(2^0+2^1+2^2+2^3+...+2^(2010)`

`A=2^(2011)-1`

`A=B`

Ta có : \(\frac{2^9}{3^{2010}}:\frac{3^9}{2^{2010}}=\frac{2^{2019}}{3^{2019}}=\left(\frac{2}{3}\right)^{2019}< 1^{2019}=1\)

Vì \(\frac{2^9}{3^{2010}}:\frac{3^9}{2^{2010}}< 1\)

=> \(\frac{2^9}{3^{2010}}< \frac{3^9}{2^{2010}}\)

       Bài làm :

Cách 1:

Ta có :

 \(\frac{2^9}{3^{2010}}\div\frac{3^9}{2^{2010}}=\frac{2^9.2^{2010}}{3^{2010}.3^9}=\frac{2^{2019}}{3^{2019}}=\left(\frac{2}{3}\right)^{2019}< 1\)

 \(\Rightarrow\frac{2^9}{3^{2010}}< \frac{3^9}{2^{2010}}\)

Cách 2 :

Nhận thấy :

\(\Rightarrow\frac{2^9}{3^{2010}}< \frac{3^9}{2^{2010}}\)