\(5x^2-6xy+y^2=\left(9x^2-6xy+y^2\right)-4x^2=\left(3x-y\right)^2-4x^2=\left(3x-y-2x\right)\left(3x-y+2x\right)=\left(x-y\right)\left(5x-y\right)\)

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

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

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

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

\(=2x\left(x-3y\right)+5\left(x-3y\right)=\left(x-2y\right)\left(2x+5\right)\)

(x−2y)(2x+5)

Bạn giải thích câu d được ko

=(x²+6xy+9) -y²

= (x+3)²-y²

=(x+3-y)(x+3+y)

sai rồi vì (x²+6xy+9) ko có y² nên ko thể có hằng đẳng thức được 

\(a)\)\(3x^2-6xy+3y^2-12\)

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

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

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

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

\(b)\)\(x^2+5x+6\)

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

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

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

Chúc bạn học tốt ~ 

a) 3x²-6xy+3y²-12=3(x²-2xy+y²)-12=3(x-y)²-12=3[(x-y)²-4]=3(x-y-2)(x-y+2)

b)x²+3x+2x+6=x(x+3)+2(x+3)=(x+3)(x+2)

a: x^2+4xy-21y^2

\(=x^2+7xy-3xy-21y^2\)

\(=x\left(x+7y\right)-3y\left(x+7y\right)\)

\(=\left(x+7y\right)\left(x-3y\right)\)

b: \(5x^2+6xy+y^2\)

\(=5x^2+5xy+xy+y^2\)

=5x(x+y)+y(x+y)

=(x+y)(5x+y)

c: \(x^2+2xy-15y^2\)

\(=x^2+5xy-3xy-15y^2\)

=x(x+5y)-3y(x+5y)

=(x+5y)(x-3y)

d: \(x^2-7xy+10y^2\)

\(=x^2-2xy-5xy+10y^2\)

=x(x-2y)-5y(x-2y)

=(x-2y)(x-5y)

a) \(x^2+4xy-21y^2\)

\(=x^2+7xy-3xy-21y^2\)

\(=x\left(x+7y\right)-3y\left(x+7y\right)\)

\(=\left(x+7y\right)\left(x-3y\right)\)

b) \(5x^2+6xy+y^2\)

\(=5x^2+5xy+xy+y^2\)

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

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

c) \(x^2+2xy-15y^2\)

\(=x^2+5xy-3xy-15y^2\)

\(=x\left(x+5y\right)-3y\left(x+5y\right)\)

\(=\left(x+5y\right)\left(x-3y\right)\)

d) \(x^2-7xy+10y^2\)

\(=x^2-2xy-5xy+10y^2\)

\(=x\left(x-2y\right)-5y\left(x-2y\right)\)

\(=\left(x-5y\right)\left(x-2y\right)\)

`1)`

`a)3x^2-6xy+3y^2=3(x^2-2xy+y^2)=3(x-y)^2`

`b)(x-y)^2-4x^2=(x-y-2x)(x-y+2x)=(-x-y)(3x-y)`

`2)`

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

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

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

`<=>[(x=3),(x=1/2):}`

`b)x^2+5x+6=0`

`<=>x^2+2x+3x+6=0`

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

`<=>[(x=-2),(x=-3):}`

C

5x² + 6xy + y²

= 5x² + 5xy + xy + y²

= ( 5x² + 5xy ) + ( xy + y² )

= 5x( x + y ) + y( x + y )

= ( x + y )( 5x + y )

\(5x^2+6xy+y^2\)   

\(=5x^2+5xy+xy+y^2\)   

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

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