(2$x^{2}$ +2x+5) chia hết cho x+2

Đáp án:

Giải thích các bước giải:

 `2x^2+2x+5 vdots x+2`
 `2x*x+2x+5 vdots x+2`

 `2x(x+2-2)+(2x+4)+1 vdots x+2`

 `2x(x+2) - 4x + 2(x + 2) + 1 vdots x+2`

 `2x(x+2) + 2(x + 2)-4x - 8+9 vdots x+2`

 `2x(x+2) + 2(x+2) -(4x+8)+9 vdots x+2`

`2x(x+2) + 2(x+2) - 4(x+2) + 9 vdots x+2`

Mà `2x(x+2); 2(x+2); 4(x+2) vdots x+2`

`=> 9 vdots x+2`
`=> x + 2 in Ư(9)`

`=> x+2 in {-9;-3;-1;1;3;9}`

Ta có bảng sau:

|__`x+2`__|_`-9`__|_`-3`_|_`-1`_|__`1`__|__`3`__|__`9`__|

|____`x`____|_`-11`_|_`-5`_|_`-3`_|_`-1`_|__`1`__|__`7`__|

Vậy `x in {-11;-5;-3;-1;1;7}`