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

1/2²<1/1*2=1/1-1/2

1/3²<1/2*3=1/2-1/3

1/4²<1/3*4=1/3-1/4

1/2010²<1/2009*2010=1/2009-1/2010

1/2²+1/3²+1/4²+...+1/2010²<1/1-1/2+1/2-1/3+1/3-1/4+...+1/2009-1/2010

1/2²+1/3²+1/4²+...+1/2010<1/1-1/2010<1 (dfcm)

k mình nha 

Ta có: n < 1/1.2 + 1/2.3 + 1/3.4 +...+ 1/2008.2009 + 1/2009.2010

          n < 1/1-1/2 + 1/2-1/3 + 1/3-1/4 +...+ 1/2008-1/2009 + 1/2009-1/2010 (công thức)

          n < 1/1- (1/2-1/2)- (1/3-1/3)-...- (1/2009-1/2009)-1/2010 (quy tắc dấu ngoặc)

          n < 1/1 - 1/2010

          n < 2009/2010

Vậy n<2009/2010<1

ta có \(N=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2010^2}.\)

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

đặt \(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2009.2010}\)

\(\Rightarrow N< A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2009.2010}\)

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

                 \(=1-\frac{1}{2010}< 1\)

hay \(N< 1\left(đpcm\right)\)

a) A= 1/2010+1+2/2009+1+3/2008+1+...+2009/2+1+1

  = 2011/2010+20011/2009+2011/2008+...+2011/2+2011/2011

  = 2011(1/2+1/3+1/4+...+1/2011)

Ta có: B= 1/2+1/3+1/4+...+1/2011

suy ra A/B= 2011

=1/2010

\(\frac{A}{B}\)=2011

Ta có 

1/2^2<1/1.2

1/3^2<1/2.3

......

1/2009^2<1/2008.2009

1/2010^2<1/2009.2010

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

=>N<1/1.2+1/2.3+....+1/2009.2010

=>N<1-1/2010

=>N<2009/2010<1

Vậy N<1

\(N=\) \(\frac{1}{2^2}\) \(+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2009^2}+\frac{1}{2010^2}\)\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2008.2009}+\frac{1}{2009.2010}\)

\(N< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2008}-\frac{1}{2009}+\frac{1}{2009}-\frac{1}{2010}\)

\(N< 1-\frac{1}{2010}\)       

\(N< \frac{2009}{2010}< 1\)

\(\Rightarrow N< 1\)

Ta có :

\(N=\frac{2009^{2010}-2}{2009^{2011}-2}< \frac{2009^{2010}-2+2011}{2009^{2011}-2+2011}\)

\(=\frac{2009^{2010}+2009}{2009^{2011}+2009}=\frac{2009.\left(2009^{2009}+1\right)}{2009.\left(2009^{2010}+1\right)}\)

\(=\frac{2009^{2009}+1}{2009^{2010}+1}=M\)

Vậy \(M>N\)

Ta có: \(B< 1\)

\(\Rightarrow B< \frac{2009^{2010}-2+3}{2009^{2011}-2+3}=\frac{2009^{2010}+1}{2009^{2011}+1}\left(1\right)\)

Mà \(\frac{2009^{2010}+1}{2009^{2011}+1}< 1\)

\(\Rightarrow\frac{2009^{2010}+1}{2009^{2011}+1}< \frac{2009^{2010}+1+2008}{2009^{2011}+1+2008}=\frac{2009^{2010}+2009}{2009^{2011}+2009}=\frac{2009\left(2009^{2009}+1\right)}{2009\left(2009^{2010}+1\right)}=\frac{2009^{2009}+1}{2009^{2010}+1}=A\left(2\right)\)

Từ (1) và (2) suy ra A > B

Sửa A vs B thành M vs N nhé quen ghi A vs B nên....