a) Nhóm x^2 và y^2  ; x và y 

b) Nhóm 3 hạng tử đầu lại vs nhau . Sau cùng xuất  hiện nhân tử chung là 3

c) Nhóm 2 hạng tử đầu với nhau. ba hạng tử còn lại với nhau . 

d) .....

D,ghép đầu với cuối là hằng dẳng thức 2 cái giữa với nhau là nhân tử chung là 3x

a) x²+y²-4x+4y+8=0

⇔ (x-2)²+(y+2)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-2\end{matrix}\right.\)

b)5x²-4xy+y²=0

⇔ x²+(2x-y)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\2x-y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)

c)x²+2y²+z²-2xy-2y-4z+5=0

⇔ (x-y)²+(y-1)²+(z-2)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y-1=0\\z-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y=1\\z=2\end{matrix}\right.\)

b: Ta có: \(5x^2-4xy+y^2=0\)

\(\Leftrightarrow x^2-\dfrac{4}{5}xy+y^2=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{2}{5}y+\dfrac{4}{25}y^2+\dfrac{21}{25}y^2=0\)

\(\Leftrightarrow\left(x-\dfrac{2}{5}y\right)^2+\dfrac{21}{25}y^2=0\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)

d)3x²+3y²+3xy-3x+3y+3=0

⇔ 6x²+6y²+6xy-6x+6y+6=0

⇔ 3(x+y)²+3(x-1)²+3(y+1)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\x-1=0\\y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)

1/x^3 - 2x^2 - 9x + 18

= x\(^2\)( x - 2 ) - 9 ( x - 2 ) = ( x\(^2\) - 9 ) ( x - 2 )= ( x - 3 ) ( x +3 ) ( x - 2 )

2/3x^2 -5x - 3y^2 + 5y

= 3( x\(^2\) - y\(^2\) ) - 5 ( x - y ) = 3 ( x - y ) ( x + y ) - 5 ( x - y ) = ( x - y ) [ 3( x+ y ) - 5 ]

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

3/49 - x^2 + 2xy - y^2

= 49 - ( x\(^2\) - 2xy + y\(^2\) ) = 49 - ( x - y )\(^2\) = ( 7 - x + y ) ( 7 + x - y )

5/ x^2 - 4x^2y^2 + 2xy

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

6/ 3x - 3y - x^2 + 2xy - y^2

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

b)x²+2xy+y²-16=(x+y)²-4²=(x+y+4)(x+y-4)

c)3x²+5x-3xy-5y=x(3x+5)-y(3x+5)=(3x+5)(x-y)

d)4x²-6x³y-2x²+8x=2x(2x-3x²y-x+4)

e)x²-4-2xy+y²=(x²-2xy+y²)-4=(x-y)²-2²=(x-y-2)(x-y+2)

k)x²-y²-z²-2yz=x²-(y+z)²=(x-y-z)(x+y+z)

m)6xy+5x-5y-3x²-3y²=3(x²-2xy+y²)+5(x-y)=3(x-y)²+5(x-y)=(x-y)(3x-3y+5)

b. (x^2+2xy+y^2)-16 =(x+y)^2-16=(x+y+4)(x+y-4)

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

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

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

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

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

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

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

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

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

=3(x-y)-(x-y)^2

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