Ta có : \(\frac{1}{2^2}< \frac{1}{1\cdot2}\)

           \(\frac{1}{3^2}< \frac{1}{2\cdot3}\)

             \(.\)                   \(.\)

             \(.\)

             \(.\)                    \(.\)  

             \(.\)                    \(.\)

         \(\frac{1}{2013^2}< \frac{1}{2012\cdot2013}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+.........+\frac{1}{2013^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+.....+\frac{1}{2012\cdot2013}\)

Mà \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+.....+\frac{1}{2012\cdot2013}=1-\frac{1}{2013}< 1\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+......+\frac{1}{2013^2}< 1\)

Nhớ k cho mình nhé!

Chúc các bạn học tốt!

mình giải ở đè trước rồi

Đặt \(A=\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}\)

\(\Rightarrow A< \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

​\(\Rightarrow A< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)​

\(\Rightarrow A< \frac{1}{2}-\frac{1}{100}< \frac{1}{2}\)

Vậy \(A< \frac{1}{2}\left(đpcm\right)\)

Ta có: \(\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}>\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

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

Help me

Bài này mình chịu thôi. Nhường cơ hội cho các bạn khác đi.

a=5   

b=4

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left[x+1\right]}=\frac{2007}{2009}\)

\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left[x+1\right]}=\frac{2007}{2009}\)

\(2\left[\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\left[x+1\right]}\right]=\frac{2007}{2009}\)

\(2\left[\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right]=\frac{2007}{2009}\)

\(2\left[\frac{1}{2}-\frac{1}{x+1}\right]=\frac{2007}{2009}\)

\(1-\frac{2}{x+1}=\frac{2007}{2009}\)

\(\frac{2}{x+1}=1-\frac{2007}{2009}\)

\(\frac{2}{x+1}=\frac{2}{2009}\)

\(\Rightarrow x+1=2009\Leftrightarrow x=2008\)

Lời giải:

Ta có:

\(\frac{1}{2^2}=\frac{1}{2.2}>\frac{1}{2.3}\)

\(\frac{1}{3^2}=\frac{1}{3.3}>\frac{1}{3.4}\)

.........

\(\frac{1}{2012^2}=\frac{1}{2012.2012}>\frac{1}{2012.2013}\)

Cộng theo vế ta có:

\(B>\underbrace{\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{2012.2013}}_{M}(1)\)

\(M=\frac{3-2}{2.3}+\frac{4-3}{3.4}+....+\frac{2013-2012}{2012.2013}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2012}-\frac{1}{2013}\)

\(=\frac{1}{2}-\frac{1}{2013}(2)\)

Từ \((1);(2)\Rightarrow B>\frac{1}{2}-\frac{1}{2013}(*)\)

---------------------------

\(B=\frac{1}{2^2}+\frac{3^2}+\frac{1}{4^2}+....+\frac{1}{2012^2}<\underbrace{ \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2011.2012}}_{N}(3)\)

Mà:

\(N=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+....+\frac{2012-2011}{2011.2012}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2011}-\frac{1}{2012}\)

\(=1-\frac{1}{2012}<1(4)\)

Từ \((3);(4)\Rightarrow B< N< 1(**)\)

Từ \((*); (**)\) ta có đpcm.

\(\frac{1}{9}+\frac{8}{9}=\frac{1+8}{9}=\frac{9}{9}=1\)

\(\frac{1}{12}+\frac{2}{12}+\frac{6}{12}+\frac{3}{12}=\frac{1+2+6+3}{12}=\frac{12}{12}=1\)

Chúc bạn học tốt !!

\(\frac{1}{9}\)+\(\frac{8}{9}\)=\(\frac{1+8}{9}\)=\(\frac{9}{9}\)=\(1\)

\(\frac{1}{12}\)+\(\frac{2}{12}\)+\(\frac{6}{12}\)+\(\frac{3}{12}\)=\(\frac{1+2+6+3}{12}\)=\(\frac{12}{12}\)=\(1\)