1: \(3x\left(x-1\right)+7x^2\left(x-1\right)\)

\(=\left(x-1\right)\left(7x^2+3x\right)\)

=x(7x+3)(x-1)

2: \(9\left(x+5\right)^2-\left(x-7\right)^2\)

\(=\left(3x+15\right)^2-\left(x-7\right)^2\)

=(3x+15+x-7)(3x+15-x+7)

=(4x+8)(2x+22)

\(=4\left(x+2\right)\cdot2\cdot\left(x+11\right)=8\left(x+2\right)\left(x+11\right)\)

3: \(x^2-2xy+y^2-xz+yz\)

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

=(x-y)(x-y-z)

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

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

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

5: \(9x^2y^2+15x^2y-21xy^2\)

\(=3xy\cdot3xy+3xy\cdot5x-3xy\cdot7y\)

=3xy(3xy+5x-7y)

6: \(x^3-2x^2y+xy^2-9xy^4\)

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

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

7: \(x^2+x-12\)

\(=x^2+4x-3x-12\)

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

=(x+4)(x-3)

8: (x+1)(x+3)(x+5)(x+7)+15

\(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)

\(=\left(x^2+8x+7\right)^2+8\left(x^2+8x+7\right)+15\)

\(=\left(x^2+8x+7+3\right)\left(x^2+8x+7+5\right)\)

\(=\left(x^2+8x+10\right)\left(x^2+8x+12\right)=\left(x^2+8x+10\right)\left(x+2\right)\left(x+6\right)\)

Bài 1 :

\(x^2-6x+8=x^2-2x-4x+8=x\left(x-2\right)-4\left(x-2\right)=\left(x-4\right)\left(x-2\right)\)

Bài 2 :

 \(x^8+x^7+1=x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1-x^6-x^5-x^4-x^3-x^2-x\)

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

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

Tick đúng nha 

X^2-6+8

1) Ta có : 2x² + 3x - 5

= 2x² - 2x + 5x - 5

= 2x(x - 1) + 5(x - 1)

= (x - 1) (2x + 5) 

3) x² + x - 6

= x² + 2x - 3x - 6

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

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

= (x - 3)(x + 2)