a. \(=4x^3-12x^2-x^2+3x+6x-18=\left(x-3\right)\left(4x^2-x+6\right)\)

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

c.  \(=x^3+2x^2-6x^2-12x+4x+8=\left(x+2\right)\left(x^2-6x+4\right)\)

a) \(x^3+4x^2-21x\)

\(=x\left(x^2+4x-21\right)\)

\(=x\left(x^2-3x+7x-21\right)\)

\(=x\left[x\left(x-3\right)+7\left(x-3\right)\right]\)

\(=x\left(x-3\right)\left(x+7\right)\)

b) \(5x^3+6x^2+x\)

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

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

\(=x\left[5x\left(x+1\right)+\left(x+1\right)\right]\)

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

c) \(x^3-7x+6\)

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

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

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

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

d) \(3x^3+2x-5\)

\(=3x^3+3x^2+5x-3x^2-3x-5\)

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

\(=\left(x-1\right)\left(3x^2+3x+5\right)\)

a) \(=x^2-2x-4x+8\)

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

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

c) \(=x^3-x-6x-6\)

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

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

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

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

Làm 1 cách là đủ rồi mà (: 6 cách thì đến bao giờ :v

a) x² + x - 6 = x² - 2x + 3x - 6 = x( x - 2 ) + 3( x - 2 ) = ( x - 2 )( x + 3 )

b) x² - 4x + 3 = x² - x - 3x + 3 = x( x - 1 ) - 3( x - 1 ) = ( x - 1 )( x - 3 )

c) x² + 5x + 4 = x² + x + 4x + 4 = x( x + 1 ) + 4( x + 1 ) = ( x + 1 )( x + 4 )

d) x² - x - 6 = x² + 2x - 3x - 6 = x( x + 2 ) - 3( x + 2 ) = ( x + 2 )( x - 3 )

e) 2x² + 5x + 3 = 2x² + 2x + 3x + 3 = 2x( x + 1 ) + 3( x + 1 ) = ( x + 1 )( 2x + 3 )

g) 2x² - 7x + 3 = 2x² - 6x - x + 3 = 2x( x - 3 ) - ( x - 3 ) = ( x - 3 )( 2x - 1 )

h) 3x² + 10x - 8 = 3x² + 12x - 2x - 8 = 3x( x + 4 ) - 2( x + 4 ) = ( x + 4 )( 3x - 2 )

k) \(\frac{1}{2}x^2-\frac{19}{6}x+1=\frac{1}{2}x^2-\frac{1}{6}x-3x+1=\frac{1}{2}x\left(x-\frac{1}{3}\right)-3\left(x-\frac{1}{3}\right)=\left(x-\frac{1}{3}\right)\left(\frac{1}{2}x-3\right)\)

bị đòi làm 6 cách :')

a. 6x³-x²-486x+81

= 6x³-54x²+53x²-477x-9x+81

= 6x².(x-9)+53x.(x-9)-9.(x-9)

= (x-9).(6x²+53x-9)

= (x-9)(6x²+54x-x-9)

=(x-9)[6x.(x+9)-(x+9)]=(x-9)(x+9)(6x-1)

b. x³-5x²+3x+9

= x³+x²-6x²-6x+9x+9

=x².(x+1)-6x.(x+1)+9.(x+1)

=(x+1)(x²-6x+9)=(x+1)(x-3)²

c. x³+3x²+6x+4

= x³+x²+2x²+2x+4x+4

= x².(x+1)+2x.(x+1)+4.(x+1)

= (x+1)(x²+2x+4)

d. 

a) x² + 4x + 3

= x² + 3x + x +3

= ( x² + 3 ) + ( x + 3 )

= x ( x + 3 ) + ( x + 3 )

= ( x + 3 ) ( x + 1 )

b) 4x² - 4x - 3

= 4x² + 2x - 6x - 3

= ( 4x² + 2x ) - ( 6x + 3 )

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

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

c) x² - x - 12

= x² + 3x - 4x - 12

= ( x² + 3x ) - ( 4x + 12 )

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

= ( x + 3 ) ( x - 4 )

d) 4x⁴ - 4x²y² - 8y⁴

= 4 ( x⁴ - x²y² - 2y⁴ )

Hk tốt

cảm ơn bạn

a) \(x^2+4x+3\)

= \(x^2+x+3x+3\)

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

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

b) \(4x^2-4x-3\)

= \(4x^2+2x-6x-3\)

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

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

c)\(x^2-x-12\)

= \(x^2-4x+3x-12\)

= \(x\left(x-4\right)+3\left(x-4\right)\)

= \(\left(x-4\right)\left(x+3\right)\)

câu d la 4x^2.y^2 phai ko

\(4x^4-4x^2y^2-8y^4\)

= \(4x^4-8x^2y^2+4x^2y^2-8y^4\)

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

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

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

Kết bạn với mình nha

\(x^2-6x+8\)

\(=x^2-2x-4x+8\)

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

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

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

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

\(=x\left(x-1\right)-3\left(x-1\right)\)

\(=\left(x-3\right)\left(x-1\right)\)

\(x^2-x-12\)

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

\(=x\left(x-4\right)+3\left(x-4\right)\)

\(=\left(x+3\right)\left(x-4\right)\)

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

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

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

Tách hạng tử là gì vậy? Tớ ko biết nó là gì nên lm 2 cách

a) C1: \(x^2-6x+8\)

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

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

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

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

C2: \(x^2-6x+8\)

\(=x^2-4x-2x+8\)

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

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

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

b) \(x^2-4x+3\)

C1: \(=x^2-2.2x+2^2-1^2\)

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

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

\(=\left(x-3\right)\left(x-1\right)\)

C2: \(=x^2-x-3x+3\)

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

\(=x\left(x-1\right)-3\left(x-1\right)\)

\(=\left(x-3\right)\left(x-1\right)\)

c) \(x^2-x-12\)

C1: \(=x^2-2.\dfrac{1}{2}x+\left(\dfrac{1}{2}\right)^2-\left(\dfrac{7}{2}\right)^2\)

\(=\left(x-\dfrac{1}{2}\right)^2-\left(\dfrac{7}{2}\right)^2\)

\(=\left(x-\dfrac{1}{2}-\dfrac{7}{2}\right)\left(x-\dfrac{1}{2}+\dfrac{7}{2}\right)\)

\(=\left(x-4\right)\left(x+3\right)\)

C2: \(=x^2-4x+3x-12\)

\(=x\left(x-4\right)+3\left(x-4\right)\)

\(=\left(x+3\right)\left(x-4\right)\)

\(\)