I DON'no

X2 - 5 xy + 6 y2 = 0

me too

a: \(=4xy\left(1-5x^2y\right)\)

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

c: \(=x\left(x-a\right)+y\left(x-a\right)=\left(x-a\right)\left(x+y\right)\)

d: \(=\left(x+2y\right)^2-36=\left(x+2y+6\right)\left(x+2y-6\right)\)

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=)))))

=[(x⁴)²+2x⁴2+2²]-4x⁴

=(x⁴+2)²-(2x²)²

⁼(x⁴+2-2x²)+(x⁴+2+2x²)

Lời giải:

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

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