Hình như đề sai r

1/2^3 = 1/8 > 1/9

Đặt \(A=\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{60^2}\)

\(A< \dfrac{1}{3^2}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{59.60}\)

\(A< \dfrac{1}{3^2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{59}-\dfrac{1}{60}\)

\(A< \dfrac{1}{3^2}+\dfrac{1}{3}-\dfrac{1}{60}\)

\(A< \dfrac{4}{9}-\dfrac{1}{60}< \dfrac{4}{9}\) (đpcm)

Thank you ! 

A=(1/2-1).(1/3-1).(1/4-1)-(1/9-1).(1/10-1)

<=>A=(-1/2).(-2/3).(-3/4)-(-8/9).(-9/10)

<=>A=-6/24+72/90

<=>A=-1/4+4/5

<=>A=11/20>0

MÀ -1/9 < 0 suy ra: A>-1/9(đpcm)

mình sửa lại dòng thứ 4 nhé;

<=> A=-1/4-4/5

<=>A=-21/20

ta có: -1/9=-20/180

         -21/20=-189/180

mà -189>-20 suy ra A>-1/9

\(P=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\frac{1}{25}+...+\frac{1}{121}+\frac{1}{144}\)

\(\Rightarrow P=\frac{1}{4}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{11^2}+\frac{1}{12^2}\)

Ta có : \(P< \frac{1}{4}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}+\frac{1}{11.12}\)

\(\Rightarrow P< \frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)

\(\Rightarrow P< \frac{1}{4}+\frac{1}{2}-\frac{1}{12}\)

\(\Rightarrow P< \frac{2}{3}\left(đpcm\right)\)

\(P=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\frac{1}{25}+...+\frac{1}{121}+\frac{1}{144}\)

\(P=\frac{1}{4}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{11^2}+\frac{1}{12^2}\)

Có : \(P< \frac{1}{4}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{10.11}+\frac{1}{11.12}\)

\(\Rightarrow P< \frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)

\(\Rightarrow P< \frac{1}{4}=\frac{1}{2}-\frac{1}{12}\)

\(\Rightarrow P< \frac{2}{3}\)( đpcm )

\(C=\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{11^2}< \frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\)

=> \(C< \frac{1}{2}-\frac{1}{11}=\frac{9}{22}\)

\(C=\frac{1}{3^2}+\frac{1}{4^2}+..+\frac{1}{11^2}>\frac{1}{3.4}+\frac{1}{4.5}+..+\frac{1}{11.12}=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+..+\frac{1}{11}-\frac{1}{12}\)

\(=>C>\frac{1}{3}-\frac{1}{12}=\frac{3}{12}=\frac{1}{4}\)

=> 1/4 < C < 9/22