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)

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

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

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

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

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

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

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

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

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

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

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

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

Chúc bạn học tốt!!!

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

Câu a:

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

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

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

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

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

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

Ta có

a,    x²-x-y²-y

=x²-y²-(x+y)

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

=(x+y)(x-y-1)

b,   x²-2xy+y²-z²

=(x-y)²-z²

=(x-y-z)(x-y+z)

con bai 32, 33 neu ban tra loi duoc minh h them

nhanh nha