\(x^4+2x^3-6x-9\)  

\(=x^4-9+2x^3-6x\) 

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

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

\(5x(2x+3)+6x+9\\=5x(2x+3)+3(2x+3)\\=(2x+3)(5x+3)\)

a: \(5x\left(2x+3\right)+6x+9\)

\(=5x\left(2x+3\right)+\left(6x+9\right)\)

\(=5x\left(2x+3\right)+3\left(2x+3\right)\)

\(=\left(2x+3\right)\left(5x+3\right)\)

b: \(3x\left(x+4\right)+48\left(x+4\right)+5\left(x+4\right)\)

\(=\left(x+4\right)\left(3x+48+5\right)\)

=(x+4)(3x+53)

SORY I'M IN GRADE 6

Thợ Đào Mỏ Panda, mày bị điên à, không biết còn trả lời làm cái quái gì

Đã có đáp án:

2x^4-6x^3+x^2+6x-3

=2x^2-6x^3-3x^2-2x^2-6x-3

=2x^2(x^2-1)-6x(x^2-1)+3(x^2-1)

=(x^2-1)(2x^2-6x+3)

a) x³-2x²-x+2

=x(x²-1)+2(-x²+1)

=x(x²-1)-2(x²-1)

=(x²-1)(x-2)

b)

x²+6x-y²+9

=x²+6x+9-y²

=(x+3)²-y²

^{=(x+3-y)(x+3+y)}

    \(2x^4+x^3-6x^2+x+2\) 

= \(2x^4+4x^3-3x^3-6x^2+x+2\)

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

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

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

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

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

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

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

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

a) Áp dụng HĐT 1 thu được ( 2 x   +   y ) 2 .

b) Áp dụng HĐT 3 với A = 2x + l; B = x - l thu được

[(2x +1) + (x -1)] [(2x +1) - (x -1)] rút gọn thành 3x(x + 2).

c) Ta có: 9 - 6x +  x 2  -  y 2 = ( 3   -   x ) 2  -  y 2  = (3 - x - y)(3 -x + y).

d) Ta có: -(x + 2) + 3( x 2  - 4) = -{x + 2) + 3(x + 2)(x - 2)

= (x + 2) [-1 + 3(x - 2)] = (x + 2)(3x - 7).

1. \(x^3+2x^2-6x-27=\left(x-3\right)\left(x^2+5x+9\right)\)

2. \(9x^2+6x-4y^2-4y=\left(9x^2-4y^2\right)+\left(6x-4y\right)\)

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

3. \(12x^3+4x^2-27x-9=4x^2\left(3x+1\right)-9\left(3x+1\right)\)

\(=\left(3x+1\right)\left(x^2-\dfrac{9}{4}\right)=\left(x+\dfrac{1}{3}\right)\left(x+\dfrac{3}{2}\right)\left(x-\dfrac{3}{2}\right)\)

1) Ta có: \(x^3+2x^2-6x-27\)

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

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

2: Ta có: \(9x^2+6x-4y^2-4y\)

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

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

\(=x\left(2x^2-x-6\right)\)

\(=x\left(2x^2-4x+3x-6\right)\)

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

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

x(2x^2-x-6)

x(2x^2-4x+3x-6)

x[2x(x-2)+3(x-2)]

x(2x+3)(x-2)

2x^5-6x^4-2a^2x^3-6ax^3

 =(2x^5-2a^2x^3)-(6x^4+6ax^3)

 =2x^3(x^2-a^2)-6x^3(x+a)

 =2x^3(x-a)(x+a)-6x^3(x+a)

 =(x+a)(2x^4-2x^3a-6x^3)

 =(x+a) 2x^3 (x-a-3)