Ta có:\(\left(\frac{9}{11}-0,81\right)^{2005}\)=\(\left(\frac{9}{11}-\frac{81}{100}\right)^{2005}=\left(\frac{9}{1100}\right)^{2005}< \left(\frac{10}{1100}\right)^{2005}=\left(\frac{1}{110}\right)^{2005}\)

Mà \(\left(\frac{1}{110}\right)^{2005}< \left(\frac{1}{100}\right)^{2005}=\left[\left(\frac{1}{10}\right)^2\right]^{2005}=\left(\frac{1}{10}\right)^{4010}=\frac{1}{10^{4010}}\)

Vậy \(\left(\frac{9}{11}-0,81\right)^{2005}< \frac{1}{10^{4010}}\)

\(\left(\frac{9}{11}-0,81\right)^{2005}=\left(\frac{9}{1100}\right)^{2005}=0,00\left(81\right)^{2005}\)

\(\frac{1}{10^{4010}}=\frac{1}{100^{2005}}=\left(\frac{1}{100}\right)^{2005}=0,01^{2005}\)

Vì 0,00(81)<0,01 nên \(\left(\frac{9}{11}-0,81\right)^{2005}< \frac{1}{10^{4010}}\)

a) A = 1/2.5 + 1/5.8 + 1/8.11 + 1/11.14 + 1/14.17 + 1/17.20

=> 3A = 1/2 - 1/5 + 1/5 - .... + 1/14 - 1/17 + 1/17 - 1/20

=> 3A = 1/2 - 1/20 = 9/20

=> A = 3/20

b) 2004¹⁰ + 2004⁹ = 2004⁹(1+2004) = 2004⁹ . 2005

2005¹⁰  = 2005⁹  . 2005

Do 2005⁹ > 2004⁹ nên 2004¹⁰ + 2004⁹ < 2005¹⁰

\(A=\dfrac{5^{10}+1}{5^{11}+1}\)

=>\(5\cdot A=\dfrac{5^{11}+5}{5^{11}+1}=\dfrac{5^{11}+1+4}{5^{11}+1}=1+\dfrac{4}{5^{11}+1}\)

\(B=\dfrac{5^9+1}{5^{10}+1}\)

=>\(5B=\dfrac{5^{10}+5}{5^{10}+1}=1+\dfrac{4}{5^{10}+1}\)

\(5^{11}+1>5^{10}+1\)

=>\(\dfrac{4}{5^{11}+1}< \dfrac{4}{5^{10}+1}\)

=>\(\dfrac{4}{5^{11}+1}+1< \dfrac{4}{5^{10}+1}+1\)

=>5A<5B

=>A<B

Có : 10A = 10.(10^11-1)/10^12-1 = 10^12-10/10^12-1 

Vì : 0 < 10^12-10 < 10^12-1 => 10A < 1 (1)

10B = 10.(10^10+1)/10^11+1 = 10^11+10/10^11+1

Vì : 10^11+10 > 10^11+1 > 0 => 10B > 1 (2)

Từ (1) và (2) => 10A < 10B

=> A < B

Tk mk nha

\(A=\frac{10^{11}-1}{10^{12}-1}\)

\(B=\frac{10^{10}+1}{10^{11}+1}\)

Mà \(\frac{10^{11}-1}{10^{12}-1}< 1\); \(\frac{10^{10}+1}{10^{11}+1}< 1\)

\(\Rightarrow\)\(A,B< 1\)

Ta có:

\(10^{11}-1>10^{10}+1\); \(10^{12}-1>10^{11}+1\)

\(\Rightarrow A>B\)

Vậy A > B

\(A< \frac{\left(10^{10}-1\right)+11}{\left(10^{11}-1\right)+11}< \frac{10^{10}+10}{10^{11}+10}< \frac{10\left(10^9+1\right)}{10\left(10^{10}+1\right)}< \frac{10^9+1}{10^{10}+1}\)

\(\Rightarrow A< B\)

Vậy A<B