a) \(x^2+4x-y^2+4\)

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

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

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

b) \(x^2-2xy-4-4y\)

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

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

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

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

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

\(x^2-2xy+y^2-z^2\)

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

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

\(x^2-2xy+y^2-z^2\)

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

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

\(=\left(x-y\right)\left(x+1\right)\)

(x²-2xy+y²⁾-z²

<=>(x-y)²-z²

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

1.Phân tích thành nhân tử ( phương pháp nhóm nhiều hạng tử )

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

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

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

b. xy - 5x + 3y^2 - 15y

\(=xy+3y^2-5x-15y\)

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

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

c.2xy + 6x + y^2 + 3y

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

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

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

a) \(x^3+2x^2-xy-2y\)

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

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

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

\(=\left(x+2\right)\left(x+\sqrt{y}\right)\left(x-\sqrt{y}\right)\)

a) \(x^3+2x^2-xy-2y\)

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

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

\(=\left(x+2\right)\left(x-\sqrt{y}\right)\left(x+\sqrt{y}\right)\)

b) \(xy-5x+3y^2-15y\)

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

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

c) \(2xy+6x+y^2+3y\)

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

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

ai h minh minh h lai cho

x⁴y⁴ + x²y² + 2xy

= ( x³y³ + xy + 2 ) xy

= ( x³y³ + x²y² - x²y² - xy + 2xy + 2 ) xy

= [ x²y² ( xy + 1 ) - xy ( xy + 1 ) + 2 ( xy + 1)] xy

= ( x²y² - xy + 2 ) ( xy + 1 ) xy

\(3,x\left(x-1\right)-y\left(1-x\right)=\left(x+y\right)\left(x-1\right)\\ 4,x^3+6x^2y+12xy^2+8y^3=\left(x+2y\right)^3\\ 5,x^2-2xy+y^2-xz+yz=\left(x-y\right)^2-z\left(x-y\right)=\left(x-y-z\right)\left(x-y\right)\\ 6,x^2-y^2-x+y=\left(x-y\right)\left(x+y\right)-\left(x-y\right)=\left(x-y\right)\left(x+y-1\right)\\ 9,x^3+x^2-xy+xy+y^2+y^3\\ =x^2\left(x+1\right)+y^2\left(x+1\right)=\left(x^2+y^2\right)\left(x+1\right)\\ 10,x^2-6\left(x+3\right)-9\\ =x^2-6x-18-9\\ =x^2-6x-27=\left(x-9\right)\left(x+3\right)\)

10: \(x^2-6\left(x+3\right)-9\)

\(=x^2-6x-18-9\)

\(=x^2-6x-27\)

\(=\left(x-9\right)\left(x+3\right)\)

3)(x-1).(x-y)

x^2-x-y^2-y

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

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

b,=(x-y)^2-z^2

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

Các câu còn lại lm tương tự.....

h.trang à

\(x^2-x-y^2-y\)

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

\(=\left(x+y\right)\left(x-y-1\right)\)

\(x^2-2xy+y^2-z^2\)

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

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

\(5x+5y+ax-ay\)

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

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

Bài làm:

a) \(x^2-2xy+y^2-zx+yz\)

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

\(\left(x-y\right)\left(x-y-z\right)\)

a/ \(x^2-2xy+y^2-zx+yz.\)

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

\(=\left(x-y\right)\left(x-y-z\right)\)

c/ \(x^2-y^2-2x-2y.\)

\(=x^2-2x+1-y^2-2y-1\)

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

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

\(=\left(x-1+y+1\right)\left(x-1-y-1\right)\)

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

sử dụng hằng thành thạo = ez 

\(a,x^2-2xy+y^2-zx+yz\)

\(=\left(x-y\right)\left(x-y\right)-z\left(x-y\right)=\left(x-y\right)\left(x-y-z\right)\)

\(b,a^3-3a+3b-b^3=\left(a^3-b^3\right)-3\left(a-b\right)\)

\(=\left(a-b\right)\left(a^2+ab+b^2\right)-3\left(a-b\right)\)

\(=\left(a-b\right)\left(a^2+ab-b^2-3\right)\)

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

x²-x-y²-y=(x-y)(x+y)-(x+y)=(x+y)(x-y-1)

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

\(x^2-x-y^2-y\)

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

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

\(=\left(x+y\right)\left(x-y-1\right)\)