trừi ơi , bạn có thôi ngay cái tính đó ko ,

bạn nói kiểu này , có khi bạn cần bài toán nào , bạn đăng lên ko ai làm đâu

Ta có : 1/2^2<1/1.2

           1/3^2 < 1/2.3

          1/4^2<1/3.4

           ................

             .............

            1/2010^2<1/2009.2010

=> 1/2^2+1/3^2+1/4^2+1/5^2+.....+1/2010^2 < 1/1.2+1/2.3+1/3.4+....+1/2009.2010

=> 1/2^2+1/3^2+1/4^2+1/5^2+.....+1/2010^2 < 1-1/2009

=> 1/2^2+1/3^2+1/4^2+1/5^2+.....+1/2010^2 < 2008/2009 < 1

Vậy 1/2^2+1/3^2+1/4^2+1/5^2+.....+1/2010 < 1 

\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2010^2}<\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{2009.2010}\)

                                             \(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{2009}-\frac{1}{2010}=1-\frac{1}{2010}=\frac{2009}{2010}<1\)

Ta có :  

\(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{2010^2}< \frac{1}{2009.2010}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2010^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2009.2010}\)

\(\Rightarrow N< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2009}-\frac{1}{2010}\)

\(\Rightarrow N< 1-\frac{1}{2010}\)

\(\Rightarrow N< 1\left(đpcm\right)\)

Chúc bạn học tốt !!!! 

mọi người ơi tl nhanh nhanh nha mk đag rất cần

ta có: \(\frac{1}{2^2}=\frac{1}{2.2}<\frac{1}{1.2};\frac{1}{3^2}=\frac{1}{3.3}<\frac{1}{2.3};...;\frac{1}{2010^2}=\frac{1}{2010.2010}<\frac{1}{2009.2010}\)

\(\Rightarrow N<\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2009.2010}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+..+\frac{1}{2009}-\frac{1}{2010}=\frac{1}{1}-\frac{1}{2010}=\frac{2009}{2010}<1\)

=>N<1(đpcm)

A=(2+2²)+(2³+2⁴)+...+(2²⁰⁰⁹+2²⁰¹⁰) 

A=2(1+2)+2³(1+2)+...+2²⁰⁰⁹(1+2) 

A=2.3+2³.3+...+2²⁰⁰⁹.3

A=3(2+2³+...+2²⁰⁰⁹) chia hết cho 3 

\(A=2^1+2^2+...+2^{2010}\)

\(A=\left(2^1+2^2\right)+\left(2^3+2^4\right)+\left(2^5+2^6\right)+...+\left(2^{2009}+2^{2010}\right)\)

\(A=2\left(1+2\right)+2^3\left(1+2\right)+2^5\left(1+2\right)+...+2^{2009}\left(1+2\right)\)

\(A=2.3+2^3.3+2^5.3+...+2^{2009}.3\)

\(A=3.\left(2+2^3+2^5+...+2^{2009}\right)\)\(⋮\)\(3\)

\(\Rightarrow A⋮3\)

\(A=\left(2^1+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+\left(2^7+2^8+2^9\right)+...+\left(2^{2008}+2^{2009}+2^{2010}\right)\)

\(A=2\left(1+2+4\right)+2^4\left(1+2+4\right)+2^7\left(1+2+4\right)+...+2^{2008}\left(1+2+4\right)\)

\(A=2.7+2^4.7+2^7.7+...+2^{2008}.7\)

\(A=7.\left(2+2^4+2^7+...+2^{2008}\right)\)\(⋮\)\(7\)

\(\Rightarrow A⋮7\)

\(B=3^1+3^2+...+3^{2010}\)

\(B=\left(3^1+3^2\right)+\left(3^3+3^4\right)+\left(3^5+3^6\right)+...+\left(3^{2009}+3^{2010}\right)\)

\(B=3\left(1+3\right)+3^3\left(1+3\right)+3^5\left(1+3\right)+...+3^{2009}\left(1+3\right)\)

\(B=3.4+3.3^3+3.3^5+...+3^{2009}.4\)

\(B=4.\left(3+3^3+3^5+...+3^{2009}\right)\)\(⋮\)\(4\)

\(\Rightarrow B⋮4\)

\(B=3^1+3^2+...+3^{2010}\)

\(B=\left(3^1+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+\left(3^7+3^8+3^9\right)+...+\left(3^{2008}+3^{2009}+3^{2010}\right)\)

\(B=3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+3^7\left(1+3+3^2\right)+...+3^{2008}\left(1+3+3^2\right)\)

\(B=3.13+3^4.13+3^7.13+...+3^{2008}.13\)

\(B=13.\left(3+3^4+3^7+...+3^{2008}\right)\)\(⋮\)\(13\)

\(\Rightarrow B⋮13\)