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

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

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

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

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:

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

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

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

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

C1:x^2+4y^2+4xy-16

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

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

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

C2: x^3+x^2+y^3+xy

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

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

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

bài này ra lâu r nhưng ngứa tay nên giải luôn=)))))

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

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

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

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

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

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

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

2(x+3)-x³-3x

\(=-x^3-3x+2x+6\)

\(=-x^3-x+6\)

Đa thức này ko phân tích được nha bạn

a) Ta có: \(x^2-3x+xy-3y\)

\(=x\left(x-3\right)+y\left(x-3\right)\)

\(=\left(x-3\right)\left(x+y\right)\)

b) Ta có: \(x^3+10x^2+25x-xy^2\)

\(=x\left(x^2+10x+25-y^2\right)\)

\(=x\left(x+5-y\right)\left(x+5+y\right)\)

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

\(=4x^3-4\)

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

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

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

\(a,=-\left(x-1\right)^3\left[=\left(1-x\right)^3\right]\\ b,=\left(1-x\right)^3\)

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

b. \(=\left(1-x\right)^3\)

\(\Leftrightarrow x^3-2x^2+x^2-2x+x-2\)

\(\Leftrightarrow x^2\left(x-2\right)+x\left(x-2\right)+\left(x-2\right)\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+x+1\right)\)

`x^3-x^2-x-2`

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

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

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

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

`@` `\text {Ans}`

`\downarrow`

`x^3 - x^2 - x - 2`

`= x^3 - 2x^2 + x^2 - 2x + x - 2`

`= (x^3 - 2x^2) + (x^2 - 2x) + (x-2)`

`= x^2(x - 2) + x(x - 2) + (x-2)`

`= (x^2 + x + 1)(x-2)`