\(x^2+6x+5=x\left(x+1\right)+5\left(x+1\right)=\left(x+1\right)\left(x+5\right)\)

x² + 6x + 5 = x² + x + 5x + 5 = x(x + 1) + 5(x + 1) = (x + 5)(x + 1)

x²+6x+5=x²+5x+x+5=x(x+1)+5(x+1)=(x+5)(x+1)

x(x-5)(x+2)

(x² + 6x + 5)(x² - 10x + 21) - 20

= (x² + x + 5x + 5)(x² - 3x - 7x + 21)  - 20

= (x + 1)(x + 5)(x - 3)(x - 7) - 20

= (x² -2x - 3)(x² - 2x- 35) - 20

Đặt x² - 2x - 19 = a

=> (a + 16)(a - 16) - 20 = a² - 256 - 20 = a² - 276

= \(\left(a-2\sqrt{69}\right)\left(a+2\sqrt{69}\right)\)

= \(\left(a^2-2x-19-2\sqrt{69}\right)\left(x^2-2x-19+2\sqrt{69}\right)\)

C1: 

x²-6x+5=x²-x-5x+5=(x²-x)-(5x-5)=x(x-1)-5(x-1)=(x-5)(x-1)

C2: 

x²-6x+5= (x²-6x+9)-4=(x-2)²-2²=(x-2-2)(x-2+2)=x(x-4)

mới 2 cách thôi mà ;-;

\(x^2-x-12=x^2-x+\frac{1}{4}-\frac{1}{4}-12=\left(x-\frac{1}{2}\right)^2-\frac{49}{4}=\left(x+3\right)\left(x-4\right)\)

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

x^2-x-12

=x^2-x-16+4

=x^2-16-x+4

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

=(x-4)(x+3)

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+x2)2−4x(1−x2)

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

đặt \(\left(1-x^2\right)\)= a

ta có :

- a . a - 4x .a

= a ( - a - 4x )

thay a = \(\left(1+x^2\right)\) ta có

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

phân tích tiếp nhé !