b/

$\\$

`B =x^3 -19x -30`

`=> B = x^3 -5x^2 +5x^2- 25x + 6x-30`

`=> B = x^2(x-5) + 5x(x-5) + (x-5) + 6(x-5)`

`=> B = (x-5)(x^2+5x+6)`

`=> B = (x-5)[ (x^2+2x) + (3x+6)]`

`=> B = (x-5)[x(x+2)+3(x+2)]` 

`=> B =(x-5)(x+2)(x+3)`

$\\$

c/ 

Đặt `t = x^2 +x -6`  

Khi đó `C = (t +1)(t-1)+1`

`=> C = t^2 -1+1 = t^2`

Khi đó thay ngược `t = x^2+x-6` ta được :

`=> C= (x^2+x-6)^2`

`=> C =( x^2 -2x+3x -6)^2`

`=> C = [x(x-2)+3(x-2)]^2`

`=> C =(x-2)^2(x+3)^2`

 $\\$

d/

Đặt `a+b =t` 

Khi đó : `D = (t+1)^2 + (t-1)^3 -4t^2` 

`=> D = (t^2+2t+1) + (t^3 - 3t^2 + 3t -1) -4t^2`

`=> D = t^3 + (t^2 -4t^2 -3t^2) + (2t +3t) + (1-1)` 

`=> D = t^3 -6t^2 +5t `

`=> D = t(t^2 - 6t+5)`

`=> D = t(t^2 - t - 5t +5)`

`=> D = t[ t(t-1) -5(t-1)]`

`=> D = t(t-1)(t-5)`

Thay `t =a+b` ta được :

`D = (a+b)(a+b-1)(a+b-5)`    

 

 

Giải thích các bước giải:

`b)`
`B=x^3-19x-30`
`B=x^3-4x-15x-30`
`B=(x^3-4x)-(15x+30)`
`B=x.(x^2-4)-15.(x+2)`
`B=x.(x-2).(x+2)-15.(x+2)`
`B=(x+2).[x.(x-2)-15]`
`B=(x+2).(x^2-2x-15)`
`B=(x+2).(x^2+3x-5x-15)`
`B=(x+2).[x.(x+3)-5.(x+3)]`
`B=(x+2).(x+3).(x-5)`
`c)`
`C=(x^2+x-5)(x^2+x-7)+1`
Đặt `a=x^2+x-5` ta có:
`C=a.(a-2)+1`
`C=a^2-2a+1`
`C=(a-1)^2`
Thay `a=x^2+x-5` vào `C=(a-1)^2` ta có:
`C=(x^2+x-5-1)^2`
`C=(x^2+x-6)^2`
`C=(x^2+3x-2x-6)^2`
`C=[x.(x+3)-2.(x+3)]^2`
`C=(x-2)^2 . (x+3)^2`
`d)`
`D=(a+b+1)^2+(a+b-1)^3-4.(a+b)^2`
Đặt `x=a+b` ta có:
`D=(x+1)^2+(x-1)^3-4.x^2`
`D=x^2+2x+1+x^3-3x^2+3x-1-4x^2`
`D=x^3-6x^2+5x`
`D=x.(x^2-6x+5)`
`D=x.(x^2-x-5x+5)`
`D=x.[x.(x-1)-5.(x-1)]`
`D=x.(x-1).(x-5)`
Thay `x=a+b` vào `D=x.(x-1).(x-5)` ta có:
`D=(a+b).(a+b-1).(a+b-5)`