a) \(x^2-xy+x-y\)

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

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

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

b) \(x^2+2xy-4x-8y\)

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

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

c) \(x^3-x^2-x+1\)

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

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

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

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) \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-15\left(1\right)=\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]-15=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-15\)

Đặt \(t=x^2+5x+4\)

(1) trở thành: \(t\left(t+2\right)-15=t^2+2t+1-16=\left(t+1\right)^2-4^2=\left(t-3\right)\left(t+5\right)\)

Thay t: \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-15=\left(x^2+5x+4-3\right)\left(x^2+5x+4+5\right)=\left(x^2+5x+1\right)\left(x^2+5x+9\right)\)

b) \(\left(2x+5\right)^2-\left(x-9\right)^2=\left(2x+5-x+9\right)\left(2x+5+x-9\right)=\left(x+14\right)\left(3x-4\right)\)

a: Ta có: \(\left(x+1\right)\cdot\left(x+2\right)\left(x+3\right)\left(x+4\right)-15\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-15\)

\(=\left(x^2+5x\right)^2+10\left(x^2+5x\right)+24-15\)

\(=\left(x^2+5x\right)^2+10\left(x^2+5x\right)+9\)

\(=\left(x^2+5x+1\right)\left(x^2+5x+9\right)\)

b: \(\left(2x+5\right)^2-\left(x-9\right)^2\)

\(=\left(2x+5-x+9\right)\left(2x+5+x-9\right)\)

\(=\left(x+15\right)\left(3x-4\right)\)

`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)`

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) $=x\left({}^{}+2xy+{}^{}-9\right)$

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

\(a,\left(x-1\right)^2-2^2=\left(x-1-2\right)\left(x-1+2\right)=\left(x-3\right)\left(x+1\right)\\ b,=\left(2x\right)^2+2.2x.3+3^2\\ =\left(2x+3\right)^2\\ c,=x^3-\left(2y\right)^3\\ =\left(x-2y\right)\left(x^2+2xy+4y^2\right)\\ d,=x^3\left(x^2-1\right)-\left(x^2-1\right)\\ =\left(x^3-1\right)\left(x^2-1\right)\\ =\left(x-1\right)\left(x^2+x+1\right)\left(x-1\right)\left(x+1\right)\\ =\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)\)

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

\(f,=\left(2x\right)^3+3.\left(2x\right)^2.1+3.2x.1^2+1^3\\ =\left(2x+1\right)^3\)

`a, 4x^3 - 16x = 4x(x^2-4) = 4x(x-2)(x+2)`

`b, x^4 - y^4 = (x^2-y^2)(x^2+y^2) = (x-y)(x+y)(x^2+y^2)`

`c, xy^2 + x^2y + 1/4y^3`

`= y(xy + x^2 + 1/4y^2)`

`d, x^2 + 2x - y^2 + 1 = (x+1)^2 - y^2`

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

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

b) \(=\left(x^2+2x\right)+\left(10x+20\right)=x\left(x+2\right)+10\left(x+2\right)=\left(x+2\right)\left(x+10\right)\)

c) đặt \(x^2+x+1=t\)

\(\left(x^2+x+1\right)\left(x^2+x+4\right)+2=t\left(t+3\right)+2=t^2+3t+2=\left(t^2+t\right)+\left(2t+2\right)=t\left(t+1\right)+2\left(t+1\right)=\left(t+1\right)\left(t+2\right)=\left(x^2+x+2\right)\left(x^2+x+3\right)\)

Em xem gõ lại đề cho đúng nha