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

a: \(x\left(2x-y\right)-y\left(2x-y\right)=\left(2x-y\right)\left(x-y\right)\)

c: \(x^2-3x+3y-y^2\)

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

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

b: \(x^2-6x-7=\left(x-7\right)\left(x+1\right)\)

a) \(x\left(2x-y\right)-y\left(2x-y\right)=\left(2x-y\right)\left(x-y\right)\)

b) \(x^2-6x-7=x\left(x-7\right)+\left(x-7\right)=\left(x-7\right)\left(x+1\right)\)

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

d) \(x^3-xy+2y-8=\left(x-2\right)\left(x^2+2x+4\right)-y\left(x-2\right)\)

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

a)x(2x - y) - y(2x - y)

=  (x - y)(2x - y)

c)x² - 3x + 3y - y² 

= x² - y² - 3x + 3y

= (x + y)(x - y) - 3(x - y)

= (x + y - 3)(x - y)

b)x² - 6x - 7

= x² + x - 7x - 7

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

= (x - 7)(x + 1)

d)x³ - xy + 2y - 8

= x³ - 8 - xy + 2y

= x³ - 2³ - y(x - 2)

= (x - 2)(x² + 2x + 4) - y (x - 2)

= (x - 2 - y)(x² + 2x + 4)

TH1: x>0

\(4x-1=\left(2\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)\)

TH2: x<0

\(4x-1=\left(2\sqrt{-x}-1\right)\left(2\sqrt{-x}+1\right)\)

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

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

= (1 - x³ ) + ( 6x - 6x² ) 

= (1 - x ).(1 + x + x²) + 6x.(1 - x)

= (1 - x).(1+x+x² + 6x)

= (1 - x).(1 + 7x +x² )

a. \(5x^2-6x+1\)

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

\(=\left(x-1\right)\left(5x-1\right)\)

a. 5x^2-6x+15x2−6x+1

=5x^2-x-5x+15x2−x−5x+1

$=\left(x-1\right)\left(5x-1\right)$

      \(x^4+6x^3+7x^2-6x+1\)

\(=x^4+6x^3+9x^2-2x^2-6x+1\)

\(=\left(x^2\right)^2+2.x^2.3x+\left(3x\right)^2-2\left(x^2+3x\right)+1\)

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

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

Chúc bạn học tốt.

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

\(=x^4+6x^3+9x^2-2x^2-6x+1\)

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

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

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