`1.`
`(x²-x+5)(x²+1)`
`=x²(x²-x+5)+1(x²-x+5)`
`=x^4-x³+5x²+x²-x+5`
`=x^4-x³+6x²-x+5`
Ta có:`x^4-x³+6x²-x+m=(x²-x+5)(x²+1)`
`⇒x^4-x³+6x²-x+m=x^4-x³+6x²-x+5`
`⇒m=5`
Vậy `m=5`
`2.`
`(2x-1)(3x+2)(3-x)`
`=[2x(3x+2)-1(3x+2)](3-x)`
`=(6x²+4x-3x-2)(3-x)`
`=(6x²+x-2)(3-x)`
`=3(6x²+x-2)-x(6x²+x-2)`
`=18x²+3x-6-6x³-x²+2x`
`=-6x³+(18x²-x²)+(3x+2x)-6`
`=-6x³+17x²+5x-6`
`3.`
`VT=(x-y)(x^4+x³y+x²y²+xy³+y^4)`
`=x(x^4+x³y+x²y²+xy³+y^4)-y(x^4+x³y+x²y²+xy³+y^4)`
`=x^5+x^4y+x³y²+x²y³+xy^4-x^4y-x³y²-x²y³-xy^4-y^5`
`=x^5+(x^4y-x^4y)+(x³y²-x³y²)+(x²y³-x²y³)+(xy^4-xy^4)-y^5`
`=x^5-y^5`
`=VP`
`⇒đpcm`
