\(a)\) Ta có : 

\(\frac{1}{100}A=\frac{100^{2009}+1}{100^{2009}+100}=\frac{100^{2009}+100}{100^{2009}+100}-\frac{99}{100^{2009}+100}=1-\frac{99}{100^{2009}+100}\)

\(\frac{1}{100}B=\frac{100^{2010}+1}{100^{2010}+100}=\frac{100^{2010}+100}{100^{2010}+100}-\frac{99}{100^{2010}+100}=1-\frac{99}{100^{2010}+100}\)

Vì \(\frac{99}{100^{2009}+100}>\frac{99}{100^{2010}+100}\) nên \(1-\frac{99}{100^{2009}+100}< 1-\frac{99}{100^{2010}+100}\)

Do đó : 

\(\frac{1}{100}A< \frac{1}{100}B\)\(\Rightarrow\)\(A< B\)

Vậy \(A< B\)

Chúc bạn học tốt ~ 

 Ta xét A= \(\frac{1}{5^2}+\frac{1}{6^2}+..+\frac{1}{100^2}\)

\(\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}>\frac{1}{5.6}+\frac{1}{6.7}...+\frac{1}{100.101}\)

=> \(A>\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{101}\)

=> \(A>\frac{1}{5}-\frac{1}{101}\)

=> \(A>\frac{96}{505}>\frac{96}{576}=\frac{1}{4}\)

Ta có : \(A< \frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)

=> \(A< \frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)

=> \(A< \frac{1}{4}-\frac{1}{100}\)

=> \(A< \frac{6}{25}< \frac{6}{24}=\frac{1}{4}\)

dễ mà k đi

a; A = 1 + 1/2^2 + 1/3^2 + 1/4^2 +...+ 1/100^2 < 2

1 = 1 = 1

1/2^2 < 1/1.2 = 1/1 - 1/2

1/3^2 < 1/2.3 = 1/2 - 1/3

.......................

1/100^2 < 1/99.100 = 1/99 - 1/100

Cộng vế với vế ta có:

A = 1 + 1 - 1/100

A = 2 - 1/100 < 2 (đpcm)

Bài 2:

M = 1/2.3/4.5/6...99/100

Ta có: \(\frac{a}{b}\) = 1 - \(\frac{b-a}{b}\) (a; b; n ∈ N* và b > a)

\(\frac{a+n}{b+n}\) = 1 - \(\frac{b-a}{b+n}\)

\(\frac{a}{b}\) < \(\frac{a+n}{b+n}\)

Áp dụng công thức trên ta có:

\(\frac12<\frac{1+1}{2+1}=\frac23\)

\(\frac34<\frac{3+1}{4+1}=\frac45\)

\(\frac56\) < \(\frac{5+1}{6+1}\) = \(\frac67\)

............................

\(\frac{99}{100}\) < \(\frac{99+1}{100+1}\) = \(\frac{100}{101}\)

Cộng vế với vế ta có:

M = \(\frac12\).\(\frac34\).\(\frac56\)...\(\frac{99}{100}\) < \(\frac23\).\(\frac45\)..\(\frac{100}{101}\) = N

M < N (đpcm)

b; M.N = \(\frac12\).\(\frac34\).\(\frac56\)...\(\frac{99}{100}\).\(\frac23\).\(\frac45\)..\(\frac{100}{101}\)

M.N = \(\frac{1.3.5\ldots99}{3.5\ldots101}\). \(\frac{2.4.6\ldots100}{2.4.6\ldots100}\)

M.N = 1/100.101

\(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+.....+\frac{1}{100^2}< \frac{1}{2}\)

\(\Leftrightarrow\frac{1}{2^2}.\left(1+\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{50^2}\right)\)

\(\Leftrightarrow\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3}............\)

\(\Leftrightarrow\frac{1}{2^2}+\frac{1}{3^2}+........+\frac{1}{50^2}< \frac{1}{1.2}+\frac{1}{2.3}+.......+\frac{1}{49.50}\)

\(\Leftrightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+.....+\frac{1}{49}-\frac{1}{50}\)

\(\Leftrightarrow1-\frac{1}{50}\)

\(\Leftrightarrow\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{50^2}\Rightarrow1+\frac{1}{2^2}+....+\frac{1}{50^2}< 1+1=2\)

\(\Leftrightarrow\frac{1}{2^2}.\left(1+\frac{1}{2^2}+....+\frac{1}{50^2}\right)< \frac{1}{2^2}.2=\frac{1}{2}\)

\(\LeftrightarrowĐPCM\)