A=1/2*2+1/3*3+1/4*4+...+1/10*10.

A>1/1*2+1/2*3+1/3*4+...+1/9*10.

A>1-1/2+1/2-1/3+...+1/9-1/10.

A>1-1/10.

A>9/10.

=>A>1/2.

Mà 1/2=66/132>65/132.

=>A>65/132.

Vậy A>65/132.

A=1/2^2+1/3^2+1/4^2+......+1/9^2+1/10^2

=1/4+1/3×3+1/4×4+.....+1/9×9+1/10×10

=>A>1/4+(1/3×4+1/4×5+...+1/9×10+1/10×11)

=>A>1/4+(1/3-1/11)

=>A>1/4+8/33

=>A>65/132( đpcm)

Xd

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

\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}< \frac{1}{2}\)

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

\(=\frac{1}{2}-\frac{1}{100}< \frac{1}{2}\left(đpcm\right)\)

Sửa đề: \(\frac{2}{5^2}<\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{9^2}<1\)

Ta có: \(\frac{1}{2^2}>\frac{1}{2\cdot3}=\frac12-\frac13\)

\(\frac{1}{3^2}>\frac{1}{3\cdot4}=\frac13-\frac14\)

...

\(\frac{1}{9^2}>\frac{1}{9\cdot10}=\frac19-\frac{1}{10}\)

Do đó: \(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{9^2}>\frac12-\frac13+\cdots+\frac19-\frac{1}{10}\)

=>\(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{9^2}>\frac12-\frac{1}{10}=\frac25\) (1)

Ta có: \(\frac{1}{2^2}<\frac{1}{1\cdot2}=1-\frac12\)

\(\frac{1}{3^2}<\frac{1}{2\cdot3}=\frac12-\frac13\)

...

\(\frac{1}{9^2}<\frac{1}{8\cdot9}=\frac18-\frac19\)

Do đó: \(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{9^2}<1-\frac12+\frac12-\frac13+\cdots+\frac18-\frac19=1-\frac19<1\) (2)

Từ (1),(2) suy ra \(\frac25<\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{9^2}<1\)