Đáp án + Giải thích các bước giải:
Bài `1` :
`A=(k-4)(k^2+4k+16)-(128+k^3)`
`=(k-4)(k^2+4k+4^2)-128-k^3`
`=k^3-4^3-128-k^3`
`=-64-128=-192`
`B=(2m+3n)(4m^2-6mn+9n^2)-(3m-2n)(9m^2+6mn+4n^2)`
`=(2m+3n)[(2m)^2-2m*3n+(3n)^2]-(3m-2n)[(3m)^2+3m*2n+(2n)^2]`
`=(2m)^3+(3n)^3-[(3m)^3 - (2n)^3]`
`=8m^3+27n^3-(27m^3-8n^3)`
`=8m^3+27n^3-27m^3+8n^3=-19m^3+35n^3`
Bài `2` :
`a)`
`VT = a^3+b^3=(a+b)(a^2-ab+b^2)`
`=(a+b)(a^2+2ab+b^2-3ab)`
`=(a+b)[(a+b)^2-3ab]`
`=(a+b)(a+b)^2 - 3ab(a+b)`
`=(a+b)^3-3ab(a+b)=VP`
`b)` `VP=(a-b)^3+3ab(a-b)`
`=a^3-3a^2b+3ab^2-b^3+3a^2b-3ab^2`
`=a^3-b^3=VT`
Bài `3` :
`a)` `(x-1)^3 + (2-x)(4+2x+x^2)+3x(x+2)=16`
`<=>x^3-3x^2+3x-1+2^3-x^3+3x^2+6x=16`
`<=>x^3-3x^2+3x-1+8-x^3+3x^2+6x=16`
`<=>9x+7=16`
`<=>9x=9`
`<=>x=1`
Vậy `S = {1}`
`b)` `(x+2)(x^2-2x+4)-x(x^2-2)=15`
`<=> x^3+8-x^3+2x=15`
`<=>8+2x=15`
`<=>2x=7`
`<=>x=7/2`
Vậy `S={7/2}`
`c)` `(x-3)^3-(x-3)(x^2+3x+9)+9(x+1)^2=15`
`<=>x^3-9x^2+27x-27-(x^3-27)+9(x^2+2x+1)=15`
`<=>x^3-9x^2+27x-27-x^3+27+9x^2+18x+9=15`
`<=>45x+9=15`
`<=>45x=6`
`<=>x=6/45 = 2/15`
Vậy `S={2/15}`.
