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Bài 1: 

a: \(x^3-6x^2+11x-6\)

\(=x^3-x^2-5x^2+5x+6x-6\)

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

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

b: \(x^3-6x^2-9x+14\)

\(=x^3-7x^2+x^2-7x-2x+14\)

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

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

c: \(x^3+6x^2+11x+6\)

\(=x^3+3x^2+3x^2+9x+2x+6\)

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

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

14 tháng 10 2020

a) x3 - 6x2 + 11x - 6

= ( x3 - 2x2 ) - ( 4x2 - 8x ) + ( 3x - 6 )

= x2( x - 2 ) - 4x( x - 2 ) + 3( x - 2 )

= ( x - 2 )( x2 - 4x + 3 )

= ( x - 2 )( x2 - x - 3x + 3 )

= ( x - 2 )[ x( x - 1 ) - 3( x - 1 ) ]

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

b) x3 - 6x2 - 9x + 14

= ( x3 - x2 ) - ( 5x2 - 5x ) - ( 14x - 14 )

= x2( x - 1 ) - 5x( x - 1 ) - 14( x - 1 )

= ( x - 1 )( x2 - 5x - 14 )

= ( x - 1 )( x2 + 2x - 7x - 14 )

= ( x - 1 )[ x( x + 2 ) - 7( x + 2 ) ]

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

c) x3 + 6x2 + 11x + 6

= ( x3 + 2x2 ) + ( 4x2 + 8x ) + ( 3x + 6 )

= x2( x + 2 ) + 4x( x + 2 ) + 3( x + 2 )

= ( x + 2 )( x2 + 4x + 3 )

= ( x + 2 )( x2 + x + 3x + 3 )

= ( x + 2 )[ x( x + 1 ) + 3( x + 1 ) ]

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

e) x6 - 9x3 + 8

Đặt t = x3

bthuc <=> t2 - 9t + 8 

            = t2 - t - 8t + 8

            = t( t - 1 ) - 8( t - 1 )

            = ( t - 1 )( t - 8 )

            = ( x3 - 1 )( x3 - 8 )

            = ( x - 1 )( x2 + x + 1 )( x - 2 )( x2 + 2x + 4 )

15 tháng 10 2016

g) (x+2)(x+3)(x+4)(x+5)-24 = \(\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\)

                                       =\(\left[x^2+7x+10\right]\left[x^2+7x+12\right]\)

đặt \(x^2+7x+10=a\)

ta có \(a\left(a+2\right)-24=a^2+2a-24\)

                                      \(=a^2+2a+1-25\)

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

                                      \(=\left(a+1-5\right)\left(a+1+5\right)\)

                                     \(=\left(a-4\right)\left(a+6\right)\)

\(\Rightarrow\) \(\left(x^2+7x+10-4\right)\left(x^2+7x+10+6\right)=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)

15 tháng 10 2016

a) = (x +5)2 - 22 = (x+5 -2)(x+5 +2) = (x+3)(x+7)

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

30 tháng 7 2018

a)   x3 -2x2 +5x-4

=x3-x2-x2+x+4x-4

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

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

b) x3-x2+x+3

=x3+x2-2x2-2x+3x+3

=x2(x+1) -2x(x+1)+3(x+1)

=(x2-2x+3)(x+1)

c) 6x3+x2+x+1

=6x3+ 3x2-2x2-x+2x+1

=6x2(x+\(\frac{1}{2}\)) - 2x(x+\(\frac{1}{2}\)) +2(x+\(\frac{1}{2}\))

=(6x2-2x+2) (x+\(\frac{1}{2}\))

=2( 3x2-x+1) (x+\(\frac{1}{2}\))

d)  4x3 + 6x2+4x+1

= 4x3+2x2+4x2+2x+2x+1

= 4x2(x+\(\frac{1}{2}\))+ 4x(x+\(\frac{1}{2}\))+2(x+\(\frac{1}{2}\))

= 2(2x2 +2x+1)( x+\(\frac{1}{2}\))

e) x6 -9x3+8

25 tháng 9 2017

Ta có : x3 - 7x + 6 

= x3 - x - 6x + 6 

= x(x2 - 1) - 6(x - 1)

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

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

= (x - 1) (x2 + x - 6) . 

CÁC Ý SAU TƯƠNG TỰ

19 tháng 2 2018

   x3 - 7x + 6 

= x3 - x - 6x + 6 

= x(x2 - 1) - 6(x - 1)

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

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

= (x - 1) (x2 + x - 6) . 

1 tháng 10 2020

1) \(x^3+2x-3\)

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

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

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

2) \(x^3-6x+4\)

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

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

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

1 tháng 10 2020

3) \(x^3-2x^2+1\)

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

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

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

4) \(x^3+5x^2-12\)

\(=\left(x^3+2x^2\right)+\left(3x^2+6x\right)-\left(6x+12\right)\)

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

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

15 tháng 11 2018

a, x^4+6x^3+11x^2+6x+1 
= x^4 + 6x^3 + 9x² + 2x² + 6x + 1 
= x^4 + 9x² + 1 + 6x^3 + 2x² + 6x 
= x^4 + 9x² + 1² + 2.x².3x + 2.x².1 + 2.3x.1
= (x² + 3x + 1)²

Mình làm được ý a nên tk 1 tk

15 tháng 10 2023

Bài 1 yêu cầu gì em?

Bài 2:

\(a,x\left(x-1\right)+5\left(x-1\right)=\left(x+5\right)\left(x-1\right)\\ b,3x\left(x+1\right)+3\left(x+1\right)=\left(3x+3\right)\left(x+1\right)=3\left(x+1\right)\left(x+1\right)=3\left(x+1\right)^2\\ c,x\left(x-3\right)+xy\left(x-3\right)=\left(x+xy\right)\left(x-3\right)=x\left(y+1\right)\left(x-3\right)\\ d,2x\left(x-2\right)-6\left(x-2\right)=\left(2x-6\right)\left(x-2\right)=2\left(x-3\right)\left(x-2\right)\)

15 tháng 10 2023

Bài 1:

a) \(3xy+6y\)

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

b) \(3x^2+9x\)

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

c) \(6x-9y^2\)

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

d) \(10xy^2-6x^2y\)

\(=2xy\left(5y-3x\right)\)

Bài 2:

a) \(x\left(x-1\right)+5\left(x-1\right)\)

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

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

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

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

\(=3\left(x+1\right)^2\) 

c) \(x\left(x-3\right)+xy\left(x-3\right)\)

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

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

d) \(2x\left(x-2\right)-6\left(x-2\right)\)

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

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