x³ - 3x²y + 3xy² - y³ - z³

= (x³ - 3x²y + 3xy² - y³) - z³

= (x - y)³ - z³

= (x - y - z)[(x - y)² + (x - y)z + z²]

= (x - y - z)(x² - 2xy + y² + xz - yz + z³)

--------------------

x² - y² + 8x + 6y + 7

= (x² + 8x + 16) - (y² - 6y + 9)

= (x + 4)² - (y - 3)²

= (x + 4 - y + 3)(x + 4 + y - 3)

= (x - y + 7)(x + y + 1)

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

\(=\left(x-y\right)^3-z^3\)

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

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

b: \(=x^2+8x+16-y^2+6y-9\)

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

=(x+4+y-3)(x+4-y+3)

=(x+y+1)(x-y+7)

a: \(x^2+4xy+y^2\)

\(=x^2+4xy+4y^2-3y^2\)

\(=\left(x+2y-y\sqrt{3}\right)\left(x+2y+y\sqrt{3}\right)\)

a) 8x²-2x-1=8x²+4x-2x-2x-1=(8x²+4x)-(2x+1)=4x.(2x+1)-(2x+1)=(2x+1).(4x-1)

b) x²-y²+10x-6y+16x=x²-y²+10x+10y-16y+16x=(x-y)(x+y)+10(x+y)-16(y+x)=(x+y)(x-y+10+16)

a)\(=3x\left(x+2y\right)\)

c)\(=\left(x-7\right)\left(x-1\right)\)

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

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

\(a,3x^2+6xy=3x\left(x+2y\right)\\ c,x^2-8x+7=\left(x^2-x\right)-\left(7x-7\right)=x\left(x-1\right)-7\left(x-1\right)=\left(x-1\right)\left(x-7\right)\\ b,x^2-2xy+3x-6y=\left(x^2+3x\right)-\left(2xy+6y\right)=x\left(x+3\right)-2y\left(x+3\right)=\left(x+3\right)\left(x-2y\right)\\ d,4x^2-y^2=\left(2x-y\right)\left(2x+y\right)\)

a) Cách 1.

Ta có 2xy + 3z + 6y + xz = (2xy + xz) + (3z + 6y)

= x(2 y + z)+3(z + 2 y) = (z + 2y)(x + 3).

Cách 2.

Ta có 2xy + 3z + 6y + xz = (2x1/ + 6y) + (3z + xz)

= 2y(x + 3) + z(3 + x) = (z + 2y)(x + 3).

b) Biến đổi được a 4   -   9 rt 3   +   a 2 -9a = (a- 9)a( a 2  +1).

c) Biến đổi được 3 x 2  + 5y - 3xy + (-5x) = (x - y)(3x - 5).

d) Biến đổi được  x 2  - (a + b)x + ab = (x- a)(x - b).

e) Ta có 4 x 2 - 4xy + y 2   –   9 t 2 =  ( 2 x   -   y ) 2   -   ( 3 t ) 2

= (2x - y - 3t )(2x - y + 31).

g) Ta có  x 3   -   3 x 2 y   +   3 xy 2   -   y 3   -   z 3

= ( x   -   y ) 3   -   z 3 = (x - y - z)( x 2   +   y 2   +   z 2  - 2xy + xz - yz).

h) Ta có x 2   -   y 2 + 8x + 6y+ 7 = ( x 2  +8x + 16) - ( y 2  - 6y+ 9)

= ( x   +   4 ) 2   - ( y - 3 ) 2  =(x-y + 7)(x + y + l).

a) 8x² - 2x - 1

=8x²+2x-4x-1

=2x.(4x+1)-(4x+1)

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

b) x² - y² + 10x - 6y + 16

=x²+10x+25-y²-6y-9

=(x+5)²-(y+3)²

=(x+5-y-3)(x+5+y+3)

=(x-y+2)(x+y+8)

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

b. \(8x^2-2x+12x-3=2x\left(4x-1\right)+3\left(4x-1\right)=\left(4x-1\right)\left(2x+3\right)\)

c. đề kiểu gì vậy? -2x-x để thành -3x à? xem lại đi nha

d. \(\left(x^2+10x+25\right)-\left(y^2+6y+9\right)=\left(x+5\right)^2-\left(y+3\right)^2=\left(x+5-y-3\right)\left(x+5+y+3\right)=\left(x-y+2\right)\left(x+y+8\right)\)

e. \(=x^4+2x^2y^2+y^4-x^2y^2=\left(x^2+y^2\right)^2-x^2y^2=\left(x^2+y^2-xy\right)\left(x^2+y^2+xy\right)\)

nhớ  L I K E

a.\(3x^2-11x+6\)

= \(3x^2-9x-2x+6\)

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

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

b\(8x^2+10x-3\)

=.\(8x^2-2x+12x-3\)

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

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

d.\(x^2-y^2+10x-6y+16\)

=\(\left(x^2+10x+25\right)-\left(y^2+6y+9\right)\)

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

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

=\(\left(x-y+2\right)\left(x+y+8\right)\)

e.\(x^4+x^2y^2+y^4\)

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

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

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

a)

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

\(1+8x^6y^3=1+\left(2x^2y\right)^3\\ =\left(1+2x^2y\right)\left(1+2x^2y+4x^4y^2\right)\)