= y-2xy\(^2\)-x\(^2\)y-y\(^3\)= y(1-2xy)-y(x\(^2\)+y\(^2\))= y(1-2xy-x\(^2\)-y\(^2\))= y(-x\(^2\)-2xy-y\(^2\)+1)

=-y(x\(^2\)+2xy+y\(^2\)-1)= -y[(x+y)\(^2\)-1] = -y(x+y-1)(x+y+1)

a: M(x)=A(x)+B(x)

=4x^4-7x^3+6x^2-5x-6-4x^4+7x^3-5x^2+5x+4

=x^2-2

b: C(x)=A(x)-B(x)

=4x^4-7x^3+6x^2-5x-6+4x^4-7x^3+5x^2-5x-4

=8x^4-14x^3+11x^2-10x-10

c: M(1)=1^2-2=-1

C(1)=8-14+11-10-10=5-20=-15

`a,` 

\(M\left(x\right)=A\left(x\right)+B\left(x\right)=\left(4x^4+6x^2-7x^3-5x-6\right)+\)`(-5x^2+7x^3+5x+4-4x^4)`

`M(x)=4x^4+6x^2-7x^3-5x-6-5x^2+7x^3+5x+4-4x^4`

`=(4x^4-4x^4)+(-7x^3+7x^3)+(6x^2-5x^2)+(-5x+5x)+(-6+4)`

`=x^2-2`

`b,`

`A(x)=B(x)+C(x)`

`-> C(x)=A(x)-B(x)`

`-> C(x)=(4x^4 + 6x^2 - 7x^3 - 5x - 6)-(-5x^2+7x^3+5x+4-4x^4)`

`C(x)=4x^4 + 6x^2 - 7x^3 - 5x - 6+5x^2-7x^3-5x-4+4x^4`

`= (4x^4+4x^4)+(-7x^3-7x^3)+(6x^2+5x^2)+(-5x-5x)+(-6-4)`

`= 8x^4-14x^3+11x^2-10x-10`

`c,`

`M(1)=1^2-2=1-2=-1`

`C(1)=8*1^4-14*1^3+11*1^2-10*1-10`

`=8-14+11-10-10=-6+11-10-10=5-10-10=-5-10=-15`

\(f\left(x\right)=-x^3+6x^2-x+a\)

Để \(f\left(x\right)\)chia hết cho \(x-1\)thì \(f\left(x\right)=\left(x-1\right)q\left(x\right)\)

Khi đó \(f\left(1\right)=0\Leftrightarrow-1+6-1+a=0\Leftrightarrow a=-4\)

Đặt \(f\left(x\right)=2x^3-3x^2+x+a\)

Ta có: phép chia \(f\left(x\right)\) cho \(x+2\) có dư là \(R=f\left(-2\right)\)

\(\Rightarrow f\left(-2\right)=2.\left(-2\right)^3-3.\left(-2\right)^2+\left(-2\right)+a\)

\(f\left(-2\right)=2.\left(-8\right)-3.4-2+a\)

\(f\left(-2\right)=-16-12-2+a\)

\(f\left(-2\right)=-20+a\)

Để \(f\left(x\right)\) chia hết cho \(x+2\) thì  \(R=0\) hay \(f\left(-2\right)=0\)

\(\Rightarrow-20+a=0\Leftrightarrow a=20\)

Để \(\left(4x^4-11x^3-2ax^2+5bx-6\right)⋮\left(x^2-2x-3\right)\) thì :

\(4x^4-11x^3-2ax^2+5bx-6=\left(x^2-2x-3\right)\cdot Q\)

\(4x^4-11x^3-2ax^2+5bx-6=\left(x^2-3x+x-3\right)\cdot Q\)

\(4x^4-11x^3-2ax^2+5bx-6=\left(x-3\right)\left(x+1\right)\cdot Q\)

Vì đẳng thức đúng với mọi x

+) Đặt x = 3 ta có :

\(4\cdot3^4-11\cdot3^3-2\cdot a\cdot3^2+5\cdot b\cdot3-6=\left(3-3\right)\left(3+1\right)\cdot Q\)

\(21-18a+15b=0\)

\(18a-15b=21\left(1\right)\)

+) Đặt x = -1 ta có :

\(4\cdot\left(-1\right)^4-11\cdot\left(-1\right)^3-2\cdot a\cdot\left(-1\right)^2+5\cdot b\cdot\left(-1\right)-6=\left(-1-3\right)\left(-1+1\right)\cdot Q\)

\(9-2a-5b=0\)

\(2a+5b=9\)

\(6a+15b=27\left(2\right)\)

Lấy (1) + (2) ta có : \(18a-15b+6a+15b=21+27\)

\(24a=48\)

\(a=2\)

\(\Rightarrow b=1\)

Vậy a = 2; b = 1

Lời giải:

$y-x^2y-2xy^2-y^3=y(1-x^2-2xy-y^2)$

$=y[1-(x^2+2xy+y^2)]=y[1-(x+y)^2]=y(1-x-y)(1+x+y)$