`***`Lời giải`***`

a)

`A(x)= -5x^3 - 2x^2 + x + 9x^3 - 2x^2 - ( x - 1 )`

`=-5x^3 - 2x^2 + x + 9x^3 - 2x^2 - x + 1 `

`=(-5x^3+ 9x^3) +(- 2x^2 - 2x^2)+ (x   - x) + 1 `

`=4x^3 -4x^2+ 1 `

`B(x)= -4x^3 - 2x^2 - 2 + 2x( 3 + x ) - 9x + 2x^3`

`=-4x^3 - 2x^2 - 2 + 6x+2x^2- 9x + 2x^3`

`=(-4x^3 + 2x^3)+(- 2x^2+2x^2 )+ (6x- 9x)- 2  `

`=-2x^3-3x- 2  `

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

`=x^3 -6x^2-2x - 4`

b)

`A(x) + B(x) - C(x)=4x^3 -4x^2+ 1-2x^3-3x- 2-(x^3 -6x^2-2x - 4)`

`=4x^3 -4x^2+ 1-2x^3-3x- 2-x^3 +6x^2+2x +4`

`=(4x^3 -x^3-2x^3)+(-4x^2+6x^2)+(-3x+2x)+(- 2  +4+ 1)`

`=x^3+2x^2-x+3`

c)

Ta có: `P(x) - C(x) = 4 - x^3`

`<=> P(x)  = 4 - x^3+C(x)`

`<=> P(x)  = 4 - x^3+x^3 -6x^2-2x - 4`

`<=> P(x)  =  (- x^3+x^3) -6x^2-2x+ (- 4+4)`

`<=> P(x)  = -6x^2-2x`

Đặt `P(x)=0`

`=>-6x^2-2x=0`

`<=>-2x(3x+1)=0`

`<=>`\(\left[ \begin{array}{l}-2x=0\\3x+1=0\end{array} \right.\)

`<=>`\(\left[ \begin{array}{l}x=0\\3x=-1\end{array} \right.\)

`<=>`\(\left[ \begin{array}{l}x=0\\x=\dfrac{-1}{3}\end{array} \right.\)

Vậy `x=0` hoặc `x=\frac{-1}{3}`

`a)`

`A(x)= -5x^3 - 2x^2 + x + 9x^3 - 2x^2 - ( x - 1 ) `

`A(x) = (-5x^3 + 9x^3) - ( -2x^2 - 2x^2 ) - (x - 1)`

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

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

`B(x)= -4x^3 - 2x^2 - 2 + 2x( 3 + x ) - 9x + 2x^3 `

`B(x)=-4x^3 - 2x^2 - 2 + 6x+2x^2- 9x + 2x^3`

`B(x)=(-4x^3 + 2x^3)+(- 2x^2+2x^2 )+ (6x- 9x)- 2`

`B(x)=-2x^3-3x- 2`

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

`C(x)= x^3 -6x^2-2x - 4`

`b)`

`A(x) + B(x) - C(x)=(4x^3 -4x^2+ 1) + (2x^3-3x- 2) - (x^3 -6x^2-2x - 4)`

`A(x) + B(x) - C(x)= 4x^3 - 4x^2 + 1 - 2x^3 - 3x - 2 - x^2 + 6x^2 + 4`

`A(x) + B(x) - C(x)=(4x^3 -x^3-2x^3)+(-4x^2+6x^2)+(-3x+2x)+(-2+4+1)`

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

`c)`

`P(x) - C(x) = 4 - x^3`

`=> P(x) = 4 - x^3 + C(x)`

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

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

`=> P(x) = 0.0 - 6x^2 - 2x`

`=> P(x) = - 6x^2 - 2x`

Ta có :

`P(x) = 0`

Nên :

`- 6x^2 - 2x = 0`

`-2x(3x + 1 ) = 0`

$\Rightarrow\begin{cases} 2x = 0\\3x + 1 = 0 \end{cases}$

$\Rightarrow\begin{cases} x = \dfrac{0}{2}\\3x = 0 - 1 \end{cases}$

$\Rightarrow\begin{cases} x =0\\3x = -1 \end{cases}$

$\Rightarrow\begin{cases} x = 0\\x = -\dfrac{1}{3} \end{cases}$

Vậy `x = 0` hoặc `x = -1/3`