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\(A=\frac{10}{27}+\frac{9}{16}\frac{11}{34}\)
Ta có: \(\frac{10}{27}< >\backslash\left(\frac{9}{16}< >\backslash\left(\frac{11}{34}< >Nên\backslash\left(A< >b\right)\right)\right)\backslash\left(B=\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+...+\frac{1}{22}\right)\)
\(B>\frac{1}{22}+\frac{1}{22}+\frac{1}{22}+...+\frac{1}{22}=11.\frac{1}{22}=\frac{1}{2}\)
Nên \(B>\frac{1}{2}\)
A=\(\frac{10^8+2}{10^8-1}=1+\frac{3}{10^8-1}\)
\(B=\frac{10^8}{10^8-3}=1+\frac{3}{10^8-3}\)
Vì\(10^8-1>10^8-3\)
\(\Rightarrow\frac{3}{10^8-1}< \frac{3}{10^8-3}\)
\(\Rightarrow1+\frac{3}{10^8-1}< 1+\frac{3}{10^8-3}\)
Vậy \(A< B\)
Ta có: 9^11+1= (9^2.10).9+1
= (...1)^10.9+1
=(...1).9+1
=(....9)+1
= ........0 \(⋮\) 10
=> 9^11+1\(⋮\) 10
Ta có: 911+1= (92.10).9+1
= (...1)10.9+1
=(...1).9+1
=(....9)+1
= ........0
=> 911+1 chia hết cho 10(ĐPCM)
Xét vế trái : \(T=\frac{1}{5}+\frac{1}{13}+\frac{1}{25}+...+\frac{1}{221}\)
Ta có : \(T< \frac{1}{5}+\frac{1}{12}+\frac{1}{24}+...+\frac{1}{220}\)
\(=\frac{1}{5}+\frac{1}{2}\left(\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\right)=\frac{1}{5}+\frac{1}{2}\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)\)
\(=\frac{1}{5}+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)\)
\(=\frac{1}{5}+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{11}\right)< \frac{1}{5}+\frac{1}{4}\Rightarrow T< \frac{9}{20}\)
= \(1+\frac{1}{3}+1+\frac{1}{9}+1+\frac{1}{27}+...+1+\frac{1}{3^{98}}\)\(\frac{1}{3^{98}}\)
\(=1.98+\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\right)\)
Đặt A=\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\)
\(\Rightarrow3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{97}}\)
\(\Rightarrow3A-A=2A=1-\frac{1}{3^{98}}\Rightarrow A=\frac{1-\frac{1}{2^{98}}}{2}< 1\)
\(\Rightarrow B=98+A< 98+1< 99< 100\)
\(\Rightarrow B< 100\)
ta có \(\frac{1}{10}+\frac{1}{11}+..+\frac{1}{19}< \frac{1}{10}+\frac{1}{10}+..+\frac{1}{10}=1\)
\(\frac{1}{20}+\frac{1}{21}+..+\frac{1}{39}< \frac{1}{20}+\frac{1}{20}+..+\frac{1}{20}=1\)
\(\frac{1}{40}+\frac{1}{41}+..+\frac{1}{79}< \frac{1}{40}+\frac{1}{40}+..+\frac{1}{40}=1\)
Vậy \(P< \frac{1}{9}+1+1+1=\frac{28}{9}\)