a>

\(\frac{1}{2^2}+\frac{1}{100^2}\)=1/4+1/10000

ta có 1/4<1/2(vì 2 đề bài muốn chứng minh tổng đó nhỏ 1 thì chúng ta phải xét xem có bao nhiêu lũy thừa hoặc sht thì ta sẽ lấy 1 : cho số số hạng )

1/100^2<1/2

=>A<1

\(B=\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-...+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{1}{4}-\dfrac{1}{100}< \dfrac{1}{4}\)

Nên B<\(\dfrac{1}{4}\)

B=\(\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}>\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{100.101}=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-...+\dfrac{1}{100}-\dfrac{1}{101}=\dfrac{1}{5}-\dfrac{1}{101}>\dfrac{1}{6}\)

Nên B>\(\dfrac{1}{6}\)

\(a)\) 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 \(A< B\)

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Trả lời

a) Đặt \(H=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)

\(\Rightarrow H< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)

\(\Leftrightarrow H< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)

\(\Leftrightarrow H< 1-\frac{1}{100}\)

\(\Leftrightarrow H< \frac{99}{100}\)

\(\Leftrightarrow A< 1+\frac{99}{100}\)

Ta thấy \(\frac{99}{100}< 1\Rightarrow A< 2\)

Vậy A<2 (đpcm)

b) Ta có: 1=1

             \(\frac{1}{2}+\frac{1}{3}< \frac{1}{2}+\frac{1}{2}=1\)

               \(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}< \frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=1\)

               \(\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+...+\frac{1}{15}< \frac{1}{8}+\frac{1}{8}+\frac{1}{8}+...+\frac{1}{8}=1\)

                \(\frac{1}{16}+\frac{1}{17}+...+\frac{1}{31}< \frac{1}{16}+\frac{1}{16}+...+\frac{1}{16}=1\)

                \(\frac{1}{32}+\frac{1}{33}+\frac{1}{34}+...+\frac{1}{63}< \frac{1}{32}+\frac{1}{33}+\frac{1}{34}+...+\frac{1}{63}=1\)

                 \(\Rightarrow B< 1+1+1+1+1+1\)

                 \(\Rightarrow B< 6\)

   Vậy B<6 (đpcm)