a) \(14x^2y-21xy^2+28x^2y^2=7xy\left(2x-3y+4xy\right)\)

b) \(3x^2-3xy-5x+5y=3x\left(x-y\right)-5\left(x-y\right)=\left(3x+5\right)\left(x-y\right)\)

Bạn xem lại đề. x hay $x^2$ nhỉ.

thiếu đăng lại.
Phân tích các đa thức sau thành nhân tử:
a,5x²  - 5xy + 7y - 7x ;
b,x²  + 2xy + x + 2y ;
c,x_²  - 6x - 9y²  + 9 ;

a; 5\(x^2\) - 5\(xy\) + 7y - 7\(x\)

= (5\(x^2\) - 5\(xy\)) + (7y - 7\(x\))

= 5\(x\)(\(x\) - y) - 7(\(x-y\))

= (\(x-y\)) (5\(x\) - 7)

a: =5x(x-y)-7(x-y)

=(x-y)(5x-7)

b: =x(x+2y)+(x+2y)

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

c; =(x-3)^2-9y^2

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

a

\(5x^2-5xy+7y-7x\\ =5x\left(x-y\right)+7\left(y-x\right)\\ =5x\left(x-y\right)-7\left(x-y\right)\\ =\left(5x-7\right)\left(x-y\right)\)

b

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

c

\(x^2-6x-9y^2+9\\ =x^2-6x+9-\left(3y\right)^2\\ =x^2-2.x.3+3^2-\left(3y\right)^2\\ =\left(x-3\right)^2-\left(3y\right)^2\\ =\left(x-3-3y\right)\left(x-3+3y\right)\)

`a, a^3 - a^2b + a - b`

`= a^2(a-b) + (a-b)`

`= (a^2+1)(a-b)`

`b, x^2 - y^2 + 2y - 1`

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

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

Vô đây xem: bài 1:phân tích đa thức thành nhân tửa)7x^3y-14x^2y+7xy^3b)3x^2-3xy-5x+5yc)x^2+7x+12giúp mình với  - Hoc24

\(a,=x\left(x-8\right)\\ b,=x\left(x-y\right)-6\left(x-y\right)=\left(x-6\right)\left(x-y\right)\\ c,=\left(x-3\right)^2-y^2=\left(x-y-3\right)\left(x+y-3\right)\\ d,=\left(x+y\right)\left(x^2-xy+y^2\right)+2\left(x+y\right)\\ =\left(x+y\right)\left(x^2-xy+y^2+2\right)\)

a: \(x^2-8x=x\left(x-8\right)\)

c: \(x^2-6x+9-y^2=\left(x-3-y\right)\left(x-3+y\right)\)

a: \(=\dfrac{2}{5}\left(xy-x-y^2+1\right)\)

\(=\dfrac{2}{5}\left[x\left(y-1\right)-\left(y-1\right)\left(y+1\right)\right]\)

\(=\dfrac{2}{5}\left(y-1\right)\left(x-y-1\right)\)

b: \(=x\left(x^2+2xy+y^2-9\right)\)

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

=25(y−1)(x−y−1)=25(y−1)(x−y−1)

b) $=x\left({}^{}+2xy+{}^{}-9\right)$

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

a: 6x-2y=2(3x-y)

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

Lời giải:
a. Không phân tích được nữa

b. $x^2(x-y)+4(y-x)=x^2(x-y)-4(x-y)=(x-y)(x^2-4)=(x-y)(x-2)(x+2)$
c. $x^3+2x^2y+xy^2-4x=x(x^2+2xy+y^2-4)$

$=x[(x^2+2xy+y^2)-4]=x[(x+y)^2-2^2]=x(x+y-2)(x+y+2)$

ko phân tích dc

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

\(a^2-6x-4b^2+9=\left(a-3\right)^2-4b^2\\ =\left(a-3-2b\right)\left(a-3+2b\right)\)