a) a+1 vdots a-1`

`=> a+1 - (a-1) vdots a-1`

`=> a+1-a+1 vdots a-1`

`=> 2 vdots a-1`

`=> a-1 ∈ Ư(2) = {1;-1;2;-2}`

`=> a ∈ {2;0;3;-1}`

b) C/m: `1/2^2 + 1/3^2 + 1/4^2 + ....+1/100^2 < 1`

Ta có: 

`1/2^2 < 1/(1.2)`

`1/3^2 < 1/(2.3)`

........

`1/100^2 < 1/(99.100)`

`=> 1/2^2 + 1/3^2 + 1/4^2 + ....+1/100^2  < 1/(1.2) + 1/(2.3) + .... 1/(99.100) (1) `

Lại có: `1/(1.2) + 1/(2.3) + .... 1/(99.100)`

`= 1 - 1/2 + 1/2 - 1/3 + .....+1/99-1/100`

`= 1-1/100 < 1 (2)`

Từ `(1;2) => 1/2^2 + 1/3^2 + 1/4^2 + ....+1/100^2  < 1`

(Chúc bạn học tốt)

Đáp án:

`a) \ \  a in {-1;0;2;3}`

`b) 1/2^2+1/3^2+1/4^2+...+1/100^2<1`

Giải thích các bước giải:

`a)`

`a+1 vdots a-1`

`to a-1+2 vdots a-1`

`to 2 vdots a-1`

`to a-1 in Ư(2)={-2;-1;1;2}`

`to a in {-1;0;2;3}`

`b)`

`1/2^2+1/3^2+1/4^2+...+1/100^2<1` 

Ta có : 

`1/2^2<1/1.2`

`1/3^2<1/2.3`

`1/4^2<1/3.4`

`...`

`1/100^2<1/99.100`

`to 1/2^2+1/3^2+1/4^2+...+1/100^2<1/1.2+1/2.3+1/3.4+...+1/99.100`

`to 1/2^2+1/3^2+1/4^2+...+1/100^2<1/1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100`

`to 1/2^2+1/3^2+1/4^2+...+1/100^2<1/1-1/100<1`

Vậy `1/2^2+1/3^2+1/4^2+...+1/100^2<1`