Đáp án:

 b `x^4 + x^2  + 1`

`= (x^4 + 2x^2 + 1) - x^2`

`= (x^2 + 1)^2 - x^2`

`= (x^2 - x + 1)(x^2 + x + 1)`

c, `(x - 1)(x - 2)(x - 3)(x - 4) - 99`

`= [(x - 1)(x - 4)].[(x - 2)(x - 3)] - 99`

`= (x^2 - x - 4x + 4)(x^2 - 2x - 3x + 6) - 99`

`= (x^2 - 5x + 4)(x^2 - 5x + 6) -99`

Đặt `x^2 - 5x + 5 = t` ta được

    `(t - 1)(t + 1) - 99`

`= t^2 - 1 - 99`

`= t^2 - 100`

`= t^2 - 10^2`

`= (t - 10)(t + 10)`

thay ngược vào ta được : 

`(x - 1)(x - 2)(x - 3)(x - 4) - 99`

`= (x^2 - 5x + 5 - 10)(x^2 - 5x + 5 + 10)`

`= (x^2 - 5x - 5)(x^2 - 5x + 15)`

d, `(x - y)^3 + (y - z)^3 + (z - x)^3`

`= (x - y + y - z)^3 - 3(x - y)(y - z)(x - y + y - z) + (z - x)^3`

`= (x - z)^3 - 3(x - y)(y - z)(x - z) - (x- z)^3`

`= -3(x - y)(y - z)(x - z)`

`= 3(x - y)(y - z)(z - x)`

Giải thích các bước giải:

Đáp án:

`b) x^4+x^2+1`

`=x^4-x+x^2+x+1`

`=x(x^3-1)+(x^2+x+1)`

`=x(x-1)(x^2+x+1)+(x^2+x+1)`

`=(x^2+x+1)[x(x-1)+1]`

`=(x^2+x+1)(x^2-x+1)` 

`c) (x-1)(x-2)(x-3)(x-4)-99`

`=[(x-1)(x-4)][(x-2)(x-3)]-99`

`=(x^2-5x+4)(x^2-5x+6)-99`

`=(x^2-5x+5-1)(x^2-5x+5+1)-99`

`=(x^2-5x+5)^2-1-99`

`=(x^2-5x+5)^2-100`

`=(x^2-5x+5-10)(x^2-5x+5+10)`

`=(x^2-5x-5)(x^2-5x+15)`

`d) (x-y)^3+(y-z)^3+(z-x)^3`

`=x^3-3x^2 y+3xy^2-y^3+y^3-3y^2 z+3yz^2-z^3+z^3-3z^2 x+3zx^2-z^3`

`=-3x^2 y+3xy^2-3y^2 z+3yz^2-3z^2 x+3zx^2`

`=3(-x^2 y+xy^2-y^2 z+yz^2-z^2 x+zx^2)`

`=3[-xy(x-y)+z(x-y)(x+y)-z^2(x-y)]`

`=3(x-y)(zx+zy-xy-z^2)`

`=3(x-y)[-x(y-z)+z(y-z)]`

`=3(x-y)(y-z)(z-x)`

Giải thích các bước giải: