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)

a.

\(=\left(x+1\right)^3-\left(3z\right)^3\)

\(=\left(x+1+3z\right)\left[\left(x+1\right)^2+3z\left(x+1\right)+9z^2\right]\)

\(=\left(x+3z+1\right)\left(x^2+2x+1+3zx+3z+9z^2\right)\)

b.

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

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

c.

\(=x^4-1+4x^2-4\)

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

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

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

a) Ta có: \(x^3+3x^2+3x+1-27z^3\)

\(=\left(x+1\right)^3-\left(3z\right)^3\)

\(=\left(x+1-3z\right)\left(x^2+2x+1+3xz+3z+9z^2\right)\)

b) Ta có: \(x^2-2xy+y^2-zx+yz\)

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

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

c) Ta có: \(x^4+4x^2-5\)

\(=x^4+4x^2+4-9\)

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

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

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

`b)x^3+y^3+z^3-3xyz`

`=x^3+3xy(x+y)+z^3-3xy(x+y)-3xyz`

`=(x+y)^3+z^3-3xy(x+y+z)`

`=(x+y+z)[(x+y)^2-z(x+y)+z^2]-3xy(x+y)`

`=(x+y+z)(x^2+2xy+y^2-zx-yz-3xy+z^2)`

`=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)`

giúp mk vs mk cần gấp T^T

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

Câu 1:

Phần a đề sai nên mk sửa lại:

a, x² + 5x - 14 = x² - 2x + 7x - 14 = x(x - 2) + 7(x - 2) = (x - 2)(x + 7)

b, xz + yz - 5(x + y) = z(x + y) - 5(x + y) = (x + y)(z - 5)

Câu 2:

x² - 4x = -4

\(\Leftrightarrow\) x² - 4x + 4 = 0

\(\Leftrightarrow\) (x - 2)² = 0

\(\Leftrightarrow\) x - 2 = 0

\(\Leftrightarrow\) x = 2

Vậy x = 2

Chúc bn học tốt!

a) (x³-x²)+(8x-8)=x(x-1)+8(x-1)=(x²+8)(x-1)

b) 8x³-8x²y+2xy²=2x(4x²-4xy+y²)

c) (x²+y²-z²)² - 4x²y²=(x²+y²-z²)² - (2xy)²=(x²+y²-z²-2xy)(x²+y²-z²+2xy)

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

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

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

`a)(x+2)^2+2(x^2-4)+(x-2)^2`

`=(x+2)^2+2(x-2)(x+2)+(x-2)^2`

`=(x+2+x-2)^2=(2x)^2=4x^2`

`b)x^2-x+1/4`

`=x^2-2.x .1/2+1/4=(x-1/2)^2`

`c)(x+y)^3-(x-y)^3`

`=(x+y-x+y)[(x+y)^2+(x+y)(x-y)+(x-y)^2]`

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

`=2y(3x^2+y^2)`

a) \(\left(x+2\right)^2+2\left(x^2-4\right)+\left(x-2\right)^2\)

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

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

b) \(x^2-x+\dfrac{1}{4}=\left(x-\dfrac{1}{2}\right)^2\)

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