\(a,=2xy\left(2y-x\right)\\ b,=x^2\left(x-4\right)+5\left(x-4\right)=\left(x^2+5\right)\left(x-4\right)\\ c,=\left(x-y\right)\left(x^2-25\right)=\left(x-y\right)\left(x-5\right)\left(x+5\right)\)

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

=(x-2)(y+2)

b: \(=\left(x+y\right)^2-4\)

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

a) $x^3-3x^2y+4x-12y$

$=(x^3-3x^2y)+(4x-12y)$

$=x^2(x-3y)+4(x-3y)$

$=(x-3y)(x^2+4)$

b) $4x^2-y^2+4y-4$

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

$=(2x)^2-(y^2-2\cdot y\cdot2+2^2)$

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

$=[2x-(y-2)][2x+(y-2)]$

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

c) $9x^2-6x-y^2+2y$

$=(9x^2-y^2)-(6x-2y)$

$=[(3x)^2-y^2]-2(3x-y)$

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

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

$\text{#}Toru$

bạn ấn ở chỗ x² cho rõ hơn nhé

\(a,=x\left(x-2\right)^2\\ b,=\left(x-y\right)^2-9=\left(x-y-3\right)\left(x-y+3\right)\\ c,=x^2\left(2x-1\right)-4\left(2x-1\right)=\left(x-2\right)\left(x+2\right)\left(2x-1\right)\\ d,=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)=\left(x-y\right)\left(x+y-5\right)\\ e,=3\left[\left(x-y\right)^2-4z^2\right]=3\left(x-y-2z\right)\left(x-y+2z\right)\\ f,=x\left[\left(x-2\right)^2-y^2\right]=x\left(x-y-2\right)\left(x+y-2\right)\\ g,=x\left[\left(x-y\right)^2-25\right]=x\left(x-y-5\right)\left(x-y+5\right)\\ h,=x^3-x-2x+2=x\left(x-1\right)\left(x+1\right)-2\left(x-1\right)\\ =\left(x-1\right)\left(x^2+x-2\right)=\left(x-1\right)^2\left(x+2\right)\\ i,=3x^2+3x-10x-10=\left(x+1\right)\left(3x-10\right)\)

a: Ta có: \(2\left(x-1\right)^3-5\left(x-1\right)^2-\left(x-1\right)\)

\(=\left(x-1\right)\left[2\left(x-1\right)^2-5\left(x-1\right)-1\right]\)

\(=\left(x-1\right)\left(2x^2-4x+2-5x+5-1\right)\)

\(=\left(x-1\right)\left(2x^2-9x+6\right)\)

b: Ta có: \(x\left(y-x\right)^3-y\left(x-y\right)^2+xy\left(x-y\right)\)

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

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

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

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

a) \(2\left(x-1\right)^3-5\left(x-1\right)^2-\left(x-1\right)=\left(x-1\right)\left[2\left(x-1\right)^2-5\left(x-1\right)-1\right]=\left(x-1\right)\left(2x^2-4x+2-5x+5-1\right)=\left(x-1\right)\left(2x^2-9x+6\right)\)

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

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

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

\(a.2\left(x-1\right)^3-5\left(x-1\right)^2-\left(x-1\right)\\ =\left(x-1\right)\left[2\left(x-1\right)^2-5\left(x-1\right)-1\right]\\ =\left(x-1\right)\left(2x^2-9x+6\right)\)

1A:

a: \(x^3+2x=x\left(x^2+2\right)\)

b: \(3x-6y=3\left(x-2y\right)\)

c: \(5\left(x+3y\right)-15x\left(x+3y\right)\)

\(=5\left(x+3y\right)\left(1-3x\right)\)

d: \(3\left(x-y\right)-5x\left(y-x\right)\)

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

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

1A. a. x(x²+2) 

b. 3(x-2y)

c. 5(x+3y)(1-3x) 

d. (x-y) (3-5x)

1B. a. 2x(2x-3)

b.xy(x²-2xy+5)

c. 2x(x+1)(x+2)

d. 2x(y-1)+2y(y-1)=2(y-1)(x-y)

1B:

a: \(4x^2-6x=2x\left(2x-3\right)\)

b: \(x^3y-2x^2y^2+5xy\)

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

`a, P = 2x(3 - x^2)`

`b, Q = 5x^2(x-3y)`

`c, R = xy(3x^2y^2 - 6y^2z + 1)`

a) \(P=6x-2x^3\)

\(P=2x\left(3+x^2\right)\)

b) \(Q=5x^3-15x^2y\)

\(Q=5x^2\left(x-3y\right)\)

c) \(R=3x^3y^3-6xy^3z+xy\)

\(R=xy\left(3x^2y^2-6y^2z+1\right)\)

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