Đặt \(A=\left(x-y+4\right)^2-\left(3x+3y-1\right)^2\)

Ta có:

\(\left(x-y+4\right)^2=x^2-xy+4x-yx+y^2-4y+4x-4y+16\)

                          \(=x^2+y^2-2xy+8x-8y+16\)

\(\left(3x+3y-1\right)^2=9x^2+9xy-3x+9xy+9y^2-3y-3x-3y+1\)

                               \(=9x^2+9y^2-6x-6y+18xy+1\)

Mình làm đến đây bạn trừ 2 kết quả cho nhau rồi sẽ ra

x² - 3x + 3y - y²

= (x² - y²) - (3x - 3y)

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

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

= x² - y² - 3x+3y = (x-y)(x+y) -3(x-y)

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

nhớ chọn cho mk nha!!!!!!

x²-y²-3x+3y

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

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

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

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

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

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

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

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

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

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

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

a, = 2 x ² - 6x +9x - 27

=(x-3).2x +9.(x-3)

=(x-3).(2x+9)

b, = 2x²-6xy+xy -3y²

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

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

Câu a:

\(x^2-y^2-x+3y-2=\left(x^2-2.x.\frac{1}{2}+\frac{1}{4}\right)-\left(y^2-2.y.\frac{3}{2}+\frac{9}{4}\right)\)

\(< =>\left(x-\frac{1}{2}\right)^2-\left(y-\frac{3}{2}\right)^2\)

\(< =>\left(x-\frac{1}{2}+y-\frac{3}{2}\right)\left(x-\frac{1}{2}-y+\frac{3}{2}\right)=\left(x+y-2\right)\left(x-y+1\right)\)