x3 – 4x2 – 12x + 27

(Nhóm để xuất hiện nhân tử chung)

= (x3 + 27) – (4x2 + 12x)

= (x3 + 33) – (4x2 + 12x)

(nhóm 1 là HĐT, nhóm 2 có 4x là nhân tử chung)

= (x + 3)(x2 – 3x + 9) – 4x(x + 3)

= (x + 3)(x2 – 3x + 9 – 4x)

= (x + 3)(x2 – 7x + 9)

a) (x - y)(x + y + 3).                    b) (x + y - 2xy)(2 + y + 2xy).

c) x 2 (x + l)( x 3  -  x 2  + 2).              d) (x – 1 - y)[ ( x   -   1 ) 2   +   ( x   -   1 ) y   +   y 2 ].

a) x⁶ + y⁶ = (x²)³ + (y²)³

= (x² + y²)(x⁴ - x²y² + y⁴)

b) x⁶ - y⁶

= (x³)² - (y³)²

= (x³ - y³)(x³ + y³)

= (x - y)(x² + xy + y²)(x + y)(x² - xy + y²)

\(a,4x^2-25-\left(2x-5\right)\left(2x+7\right)\)

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

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

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

\(b,x^3+27+\left(x+3\right)\left(x-9\right)\)

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

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

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

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

=.= hok tốt!!

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

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

\(a,=-\left(x-1\right)^3\left[=\left(1-x\right)^3\right]\\ b,=\left(1-x\right)^3\)

a. \(=-\left(x^3-3x^2+3x-1\right)=-\left(x-1\right)^3\)

b. \(=\left(1-x\right)^3\)

a) x3 + 3x2 – 3x – 9

= (x3 + 3x2) - (3x + 9)

= x2(x + 3) - 3(x + 3)

= (x + 3)(x2 - 3)

= (x + 3)(x + √3)(x - √3)

x 6 - y 6 = x 3 2 - y 3 2 = x 3 + y 3 x 3 - y 3 = x + y x 2 - x y + y x - y x 2 + x y + y 2

a) 10x³-90x⁵=10x³(1-9x²)=10x³(1-3x)(1+3x)

b)3a²-6a²b+5a-10ab=(3a²-6a²b)+(5a-10ab)=3a²(1-2b)+5a(1-b)=(3a²+5a)(1-2b)=a(3a+5)(1-2b)

c) 7a³-28a²+28a=7a(a²-4a+4)

d) 45a³-30a²+5a-500=5(9a³-6a²+a-100)