\(=6x\left(-y^2+x^2+2x+1\right)\\ =6x\left[\left(x^2+2x+1\right)-y^2\right]\\ =6x\left[\left(x+1\right)^2-y^2\right]\\ =6x\left(x+1-y\right)\left(x+1+y\right)\)

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

Ta có: \(1+6x-6x^2-x^3\)

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

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

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

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

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

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

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

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

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

=(1-x)(1+x^2 +7x)

Đây bạn ơi!

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

đặt y=x²+1

=>y²=(x²+1)²

y²=x⁴+2x²+1

đặt P(x)=x^4+6x^3+11x^2+6x+1

=x⁴+2x²+1+6x³+6x+9x²

=x⁴+2x+1+6x(x²+1)+9x²

thay y²=x⁴+2x²+1 và y=x²+1 ta được 

Q(y)=y²+6xy+9x²

=(y+3x)²

thay y=x²+1 ta được:

(x²+3x+1)²

vậy x^4+6x^3+11x^2+6x+1=(x²+3x+1)²

=6(x^2-4y^2)

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

\(6x^2-24y^2=6\left(x^2-4y^2\right)\)

\(=6\left(x-2y\right)\left(x+2y\right)\)

\(6x^2+7x+2\\ =6x^2+3x+4x+2\\ =3x\left(2x+1\right)+2\left(2x+1\right)\\ =\left(3x+2\right)\left(2x+1\right)\)

\(6x^2+7x+2\)

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

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

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

[Chorus]

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🐱 🐱 🐱 

1. \(=3x\left(2x+5\right)\)

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

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

1, = 3x.(2x + 5)

2. = (3x)² - 1² = (3x - 1).(3x +1 )

3, =(x² + 6x + 9) - y² = (x + 3)² - y² =(x + y -3 ). (x - y +3)

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