Câu 1:

$x^2+4y^2+4xy-16=[x^2+(2y)^2+2.x.2y]-16$

$=(x+2y)^2-4^2=(x+2y-4)(x+2y+4)$

Câu 2:

$x^3+x^2+y^3+xy=(x^3+y^3)+(x^2+xy)$

$=(x+y)(x^2-xy+y^2)+x(x+y)=(x+y)(x^2-xy+y^2+x)$

Câu 1:

\(x^2+4y^2+4xy-16\)

\(=\left(x+2y\right)^2-16\)

\(=\left(x+2y+4\right)\left(x+2y-4\right)\)

Câu 2:

\(x^3+x^2+y^3+xy\)

\(=\left(x^3+y^3\right)\left(x^2+xy\right)\)

\(=\left(x+y\right)\left(x^2-xy+y^2\right)+x\left(x+y\right)\)

\(=\left(x+y\right)\left(x^2-xy+y^2+x\right)\)

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

\(=x\left(x^2-y^2+2x+2y\right)\)

\(=\)\(x\left[\left(x+y\right)\left(x-y\right)+2\left(x+y\right)\right]\)

\(=x\left(x+y\right)\left(x-y+2\right)\)

x^3 - y^3 + 2x^2 + 2xy

= x [ ( x^2 - y^2 ) + ( 2x + 2y ) ]

= x [ ( x + y ) ( x - y ) + 2 ( x + y ) ]

= x ( x + y ) ( x - y + 2 )

\(=x\left(x^2-1\right)+3\left(x^2-1\right)=\left(x+3\right)\left(x^2-1\right)=\left(x+3\right)\left(x-1\right)\left(x+1\right)\)

\(=x^3+2+3x^3-6=4x^3-4=4\left(x^3-1\right)=4\left(x-1\right)\left(x^2+x+1\right)\)

= x(x^2 + 2xy + y^2 - 25z^2)

= x(x + y - 5z)(x + y + 5z)

Ta có: \(x^3-\left(y-2\right)^3+\left(y-x-2\right)^2\)

\(=\left(x-y+2\right)\left(x^2+xy-2x+y^2-4y+4\right)+\left(x-y+2\right)^2\)

\(=\left(x-y+2\right)\left(x^2+xy-2x+y^2-4y+4+x-y+2\right)\)

\(=\left(x-y+2\right)\left(x^2+y^2+6+xy-x-5y\right)\)

a) (x + 2)(x + 4).              b) 2(x + 6)(x + l).

c) 3(3x + 5)(x + l).           d) (6x -7y)(x + y).

\(x^3-x^2-14x+24\)

\(=x^3-2x^2+x^2-2x-12x+24\)

\(=x^2\left(x-2\right)+x\left(x-2\right)-12\left(x-2\right)\)

\(=\left(x-2\right)\left(x^2+x-12\right)\)

\(=\left(x-2\right)\left(x^2+4x-3x-12\right)\)

\(=\left(x-2\right)\left[x\left(x+4\right)-3\left(x+4\right)\right]\)

\(=\left(x-2\right)\left(x+4\right)\left(x-3\right)\)

Ta có:\(x^3-x^2-14x+24=\left(x^3-2x^2\right)+\left(x^2-2x\right)-\left(12x-24\right)\)

                                                    \(=x^2\left(x-2\right)+x\left(x-2\right)-12\left(x-2\right)\)

                                                    \(=\left(x-2\right)\left(x^2+x-12\right)\)

                                                    \(=\left(x-2\right)\left(x^2-3x+4x-12\right)\)

                                                    \(=\left(x-2\right)\left[x\left(x-3\right)+4\left(x-3\right)\right]\)

                                                    \(=\left(x-2\right)\left(x+4\right)\left(x-3\right)\)