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 

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

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

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

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

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

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

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

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

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

b)\(3a^2-6ab+3b^2-12c^2=3\left(a^2-2ab+b^2-4c^2\right)\)\(=3\left[\left(a-b\right)^2-\left(2c\right)^2\right]\)

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

\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)

\(A=3a^2c^2+bd+3abc+acd=\left(3a^2c^2+3abc\right)+\left(bd+acd\right)=3ac\left(ac+b\right)+d\left(b+ac\right)\\ =\left(3ac+d\right)\left(ac+b\right)\)

\(B=a^2c-a^2d-b^2d+b^2c=a^2\left(c-d\right)-b^2\left(c-d\right)=\left(a^2-b^2\right)\left(c-d\right)\\=\left(a-b\right)\left(a+b\right)\left(c-d\right)\)

\(C=8x^2+4xy-2ax-ay=\left(8x^2+4xy\right)-\left(2ax+ay\right)=4x\left(2x+y\right)-a\left(2x+y\right)\\ =\left(4x-a\right)\left(2x+y\right)\)

\(E=3a^2-6ab+3b^2-12c^2=3\left(a^2-2ab+b^2\right)-12c^2=3\left(a-b\right)^2-12c^2\\ =3\left[\left(a-b\right)^2-4c^2\right]=3\left(a-b-2c\right)\left(a-b+2c\right)\)

\(a,x^2-y^2-2x+2y=\left(x^2-y^2\right)-\left(2x-2y\right)=\left(x-y\right)\left(x+y\right)-2\left(x-y\right)=\left(x-y\right)\left(x+y-2\right).\) \(b,2x+2y-x^2-xy=2\left(x+y\right)-x\left(x+y\right)=\left(x+y\right)\left(2-x\right)\)

\(c,3a^2-6ab+3b^2-12c^2=3\left(a^2-2ab+b^2-4c^2\right)=3.\left(\left(a-b\right)^2-\left(2c\right)^2\right)\)

                                                     \(=3\left(a-b-2c\right).\left(a-b+2c\right)\)

\(d,x^2-25+y^2-2xy=\left(x^2-2xy+y^2\right)-5^2=\left(x-y\right)^2-5^2\)

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

\(e,a^2+2ab+b^2-ac-bc=\left(a+b\right)^2-c\left(a+b\right)=\left(a+b\right)\left(a+b-c\right)\)

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

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

\(h,x^2\left(x-1\right)+16\left(1-x\right)=x^2\left(x-1\right)-16\left(x-1\right)=\left(x-1\right)\left(x^2-16\right)=\)

                                                    \(=\left(x-1\right)\left(x-4\right)\left(x+4\right)\)

a) x²-2xy+y²-4

  =(x-y)²-4

   =(x-y-4)(x-y+4)

b) ab-a-3b+3

 =a(b-1)-3(b-1)

  =(a-3)(b-1)

câu ạ bạn làm có đúng không vậy 

a, x^2-2xy+y^2-4

=(x-y)^2-4

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

b, ab-a-3b+3

=a(b-1)-3(b-1)

=(b-1)(a-3)

\(3,\)Nhẩm nghiệm của đa thức trên ta đc : -1

Ta có lược đồ sau :

Phân tích thành nhân tử ta có :\(\left(x+1\right)\left(x^2-4\right)\)

a) 2x^2 -  2xy - 5x +5y

= (2x^2 - 2xy ) - ( 5x- 5 y)

=2x(x-y) - 5(x-y)

=(x- y). (2x- 5)

b)8x²  +4xy-2ax-ay

=(8x² +4xy) -(2ax+ay)

=4x(2x+y)-a(2x+y)

=(2x+y).(4x-a)

c)=x(x² -4x +4)

=x(x-2)²

d)=16- (x² -2xy +y^2)

=4^2-(x-y)^2

=(4-x+y).(4+x-y)

các câu còn lại tg tự

chúc bn hok tốt

1)

x²-y²-2x+2y

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

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

2)

2x+2y-x²-xy

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

=(2-x)(x+y)

3)

3a²-6ab+3b²-12c²

=3(a²-2ab+b²)-3(4c²)

=3(a-b)²-3(4c²)

=3[(a-b)²-4c² ]

=3(a-b-2c)(a-b+2c)

4)

x²-25+y²+2xy

=(x+y)²-25

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

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

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

4) x^2 - 25 + y^2 +2xy = x^2 + 2xy + y^2 - 25 = (x+y)^2 - 5^2 = (x+y-5)(x+y+5)

5) a^2 + 2ab +b^2-ac-bc= (a+b)^2- ac + bc = (a+b)^2 - c(a+b) = (a+b)(a+b-c)

6) x^2 - 2x - 4y^2 - 4y = (x^2 - 4y^2) - (2x+4y) = (x - 2y)(x+2y) - 2 (x+2y) = (x-2y-2)(x+2y)

7) x^2y - x^3 - 9y + 9x = x^2 (y-x) - 9(y-x) = (x^2 - 9)(y-x)= (x^2 - 3^2)(y-x) = (x-3)(x+3)(y-x)

- Xl câu 3 , 8 t hk biết lm