15 nha, nhớ nha, đừng làm thừa nha

`a)`

`A+B=x^3+3/2x+3`

`=>B=(x^3+3/2x+3)-A`

`=x^3+3/2x+3-(x^4+2x^3-5x-10)`

`=x^3+3/2x+3-x^4-2x^3+5x+10`

`=-x^4+(x^3-2x^3)+(3/2x+5x)+(3+10)`

`=-x^4-x^3+13/2x+13`

Vậy `B=-x^4-x^3+13/2x+13`

`b)`

`A-C=2x^2`

`=>C=A-2x^2`

`=x^4+2x^3-5x-10-2x^2`

`=x^4+2x^3-2x^2-5x-10`

Vậy `C=x^4+2x^3-2x^2-5x-10`

`c)`

`D=(2x-1).A`

`=>D=(2x-1)(x^4+2x^3-5x-10)`

`=2x(x^4+2x^3-5x-10)-1(x^4+2x^3-5x-10)`

`=2x^5+4x^4-10x^2-20x-x^4-2x^3+5x+10`

`=2x^5+(4x^4-x^4)-2x^3-10x^2+(-20x+5x)+10`

`=2x^5+3x^4-2x^3-10x^2-15x+10`

Vậy `D=2x^5+3x^4-2x^3-10x^2-15x+10`

`d)`

`A=(x+2).P`

Ta có: 

`A=x^4+2x^3-5x-10`

`=x^3(x+2)-5(x+2)`

`=(x+2)(x^3-5)`

Vì `A=(x+2).P` nên `(x+2)P=(x+2)(x^3-5)`

`-> P=x^3-5`

Vậy `P=x^3-5`

`e)`

Ta có: 

`A=x^4+2x^3-5x-10`

`=x(x^3-5)+2(x^3-5)`

`=(x^3-5)(x+2)`

Vì `A=(x^3-5).Q` nên `(x^3-5)Q=(x^3-5)(x+2)`

`-> Q=x+2`

Vậy có tồn tại đa thức `Q` là `Q=x+2`

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