1.  \(xy\left(a^2+2b^2\right)-ab\left(2x^2+y^2\right)\)

\(=xya^2+2xyb^2-2abx^2-aby^2\)

\(=xya^2-aby^2-2abx^2+2xyb^2\)

\(=ay\left(ax-by\right)-2bx\left(ax-by\right)\)

\(=\left(ay-2bx\right)\left(ax-by\right)\)

2. \(xy\left(a^2+2b^2\right)+ab\left(2x^2+y^2\right)\)

\(=xya^2+2xyb^2+2abx^2+aby^2\)

\(=xya^2+aby^2+2abx^2+2xyb^2\)

\(=ay\left(ax+by\right)+2bx\left(ax+by\right)\)

\(=\left(ay+2bx\right)\left(ax+by\right)\)

1. Ta có: \(3xy\left(a^2+b^2\right)+ab\left(x^2-9y^2\right)\)

\(=3xya^2+3xyb^2+abx^2+ab9y^2\)

\(=\left(3xya^2+abx^2\right)+\left(3xyb^2+ab9y^2\right)\)

\(=ax\left(3ya+bx\right)+3by\left(xb+3ya\right)\)

\(=\left(3ya+xb\right)\left(3yb+ax\right)\)

2.Check lại đề hộ mình nha:((

Câu 2 nên sủa lại đề nha

2. xy(a²+2b²)+ab(2x²+y²)

=xya²+xy2b²+ab2x²+aby²

=(xya²+aby²)+(xy2b²+ab2x²)

=ay(ax+by)+2bx(by+ax)

=(ax+by(ay+2bx)

Sửa lại đề ở câu 1: \(2ab\)chuyển thành \(2bx\)

1. \(2x^2+2b^2+2bx+2x+2b+2\)

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

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

2. \(4x^2+4x+10+6y+y^2\)

\(=\left(4x^2+4x+1\right)+\left(y^2+6y+9\right)\)

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

x²+(2a+b)xy+2aby²

=x²+2axy+bxy+2aby²

=(x²+bxy)+(2axy+2aby²)

=x(x+by)+2ay(x+by)

=(x+by)(x+2ay)

cảm ơn bn

,2x^2-12x+18+2xy-6y
= 2(x^2-6x+9) + 2y(x-3) 
= 2(x-3)^2 + 2y(x-3) 
= (x-3)(2x-6+2y) 

Giải thích: `x^2-5x+1`

`=x^2-2. 5/2x+25/4-21/4`

`=(x-5/2)^2-21/4`

`=(x-5/2-\sqrt{21}/2)(x-5/2+\sqrt{21}/2)`

`=(x-[5+\sqrt{21}]/2)(x-[5-\sqrt{21}]/2)`

a, 4\(x^3\).y + \(\dfrac{1}{2}\)yz

  =y.(4\(x^3\) + \(\dfrac{1}{2}\)z)

b, (a² + b² - 5)² - 2.(ab + 2)²

 = [a² + b² - 5  - \(\sqrt{2}\)(ab + 2) ].[ a² + b² - 5 + \(\sqrt{2}\)(ab +2)]

a) \(4x^3y+\dfrac{1}{2}yz=y\left(4x^3+\dfrac{1}{2}z\right)\)

b) \(\left(a^2+b^2-5\right)^2-2.\left(ab+2\right)^2\)

\(=\left[\left(a^2+b^2-5\right)+2\left(ab+2\right)\right]\left[\left(a^2+b^2-5\right)-2\left(ab+2\right)\right]\)

\(=\left[a^2+b^2-5+2ab+4\right]\left[a^2+b^2-5-2ab-4\right]\)

\(=\left[a^2+b^2+2ab-1\right]\left[a^2+b^2-2ab-9\right]\)

\(=\left[\left(a+b\right)^2-1\right]\left[\left(a-b\right)^2-9\right]\)

\(=\left[\left(a+b+1\right)\left(a+b-1\right)\right]\left[\left(a-b+3\right)\left(a-b-3\right)\right]\)

Mình đã làm sai, cô Hoài Thương đúng rồi.

-3x⁴y + 6x³y - 3x²y

= -3x²y( x² - 2x + 1 )

= -3x²y( x - 1 )²

-3x⁴y+6x³y-3x²y

=-3x²y(x²-2xy+1)

=-3x²y(x-1)²