a) 6x² + 7xy + 2y²

= 6x² + 4xy + 3xy + 2y²

= (6x² + 4xy) + (3xy + 2y²)

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

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

b) x² - y² + 10x - 6y + 16

= x² + 10x + 25 - y² - 6y - 9

= (x² + 10x + 25) - (y² + 6y + 9)

= (x + 5)² - (y + 3)²

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

= (x - y + 2)(x + y + 8)

c) 4x⁴ + y⁴

= 4x⁴ + 4x²y² + y⁴ - 4x²y²

= (2x² + y²)² - (2xy)²

= (2x² + y² - 2xy)(2x² + y² + 2xy)

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

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

\(=\left(3x^2-6xy\right)-\left(xy-2y^2\right)\)

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

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

\(=6x^2-9xy+2xy-3y^2\)

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

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

b: \(=\left(x-5\right)^2-9y^2\)

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

Bài 1: 

b: \(=\left(x-5\right)^2-9y^2\)

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

\(1,\\ a,=3x\left(x-3y\right)\\ b,=\left(x-5\right)^2-9y^2=\left(x-3y-5\right)\left(x+3y-5\right)\\ c,=3x\left(x-y\right)-2\left(x-y\right)=\left(3x-2\right)\left(x-y\right)\\ 2,\\ Sửa:x^2-6x+10=\left(x-3\right)^2+1\ge1>0,\forall x\)

1, =3x (2x -3y)

c, = 3x(x-y) -2(x-y)

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

2, Ta có: x² -6x+10= (x-3)² +11

Nhận xét: (x-3)² >= 0 với mọi số thực x

=> (x-3)² +1 >= 1 >0 (đpcm)

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

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

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

\(6x^2-7x+2\)

\(=6x^2-3x-4x+2\)

\(=\left(6x^2-3x\right)-\left(4x-2\right)\)

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

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

\(=6x^2-3x-4x+2=\left(2x-1\right)\left(3x-2\right)\)

2x2 + 4x + 2 – 2y2 (có nhân tử chung là 2)

= 2.(x2 + 2x + 1 – y2) (Xuất hiện x2 + 2x + 1 là hằng đẳng thức)

= 2[(x2 + 2x + 1) – y2]

= 2[(x + 1)2 – y2] (Xuất hiện hằng đẳng thức (3))

= 2(x + 1 – y)(x + 1 + y)

\(-7xy+3x^2+2y^2\\=3x^2-7xy+2y^2\\=3x^2-6xy-xy+2y^2\\=3x(x-2y)-y(x-2y)\\=(x-2y)(3x-y)\)