a)x^2-(a+b)x+ab

= x^2 - ax - bx + ab

= (x^2 - ax) - (bx - ab)

= x(x-a) - b(x-a)

= (x-b)(x-a) 

b)7x^3-3xyz-21x^2+9z

= 

c)4x+4y-x^2(x+y)

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

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

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

d) y^2+y-x^2+x

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

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

= (y-x+1) (x+y)

e)4x^2-2x-y^2-y

= [(2x)^2 - y^2] - (2x +y)

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

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

f)9x^2-25y^2-6x+10y

= 

ko biết làm

\(b,=x^4-2x^3-x^3+2x^2+3x^2-6x-3x+6\\ =\left(x-2\right)\left(x^3-x^2+3x-3\right)\\ =\left(x-2\right)\left(x-1\right)\left(x^2+3\right)\\ c,=x^4-2x^3+4x^3-8x^2+4x^2-8x+3x-6\\ =\left(x-2\right)\left(x^3+4x^2+4x+3\right)\\ =\left(x-2\right)\left(x^3+3x^2+x^2+3x+x+3\right)\\ =\left(x-2\right)\left(x+3\right)\left(x^2+x+1\right)\)

6) \(9x^3y^2+3x^2y^2=3x^2y^2\left(3x+1\right)\)

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

8) \(6x^2y+4xy^2+2xy=2xy\left(3x+2y+1\right)\)

9) \(5x^2\left(x-2y\right)-15x\left(x-2y\right)=5x\left(x-2y\right)\left(x-3\right)\)

10) \(3\left(x-y\right)-5x\left(y-x\right)=\left(x-y\right)\left(3+5x\right)\)

6) 9x³y² + 3x²y² = 3x²y²( 3x + 1 )

7) x³ + 2x² + 3x = x( x² + 2x + 3 )

8) 6x²y + 4xy² + 2xy = 2xy( 3x + 2y + 1 )

9) 5x²( x - 2y ) - 15x( x - 2y ) = 5x( x - 2y )( x - 3 )

10 3( x - y ) - 5x( y - x ) = 3( x - y ) + 5x( x - y ) = ( x - y )( 3 + 5x )

a, \(9x^3y^2+3x^2y^2=3x^2y^2\left(3x+1\right)\)

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

c, \(6x^2y+4xy^2+2xy=2xy\left(3x+2y+1\right)\)

d, \(5x^2\left(x-2y\right)-15x\left(x-2y\right)=\left(5x^2-15x\right)\left(x-2y\right)=5x\left(x-3\right)\left(x-2y\right)\)

e, \(3\left(x-y\right)-5x\left(y-x\right)=3\left(x-y\right)+5x\left(x-y\right)=\left(3+5x\right)\left(x-y\right)\)

a: x^3-7x-6

=x^3-x-6x-6

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

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

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

b: =2x^3+x^2-2x^2-x+6x+3

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

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

c: =2x^3-3x^2-2x^2+3x+2x-3

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

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

d: =2x^3+x^2+2x^2+x+2x+1

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

e: =3x^3+x^2-3x^2-x+6x+2

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

f: =27x^3-9x^2-18x^2+6x+12x-4

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

a) \(x^3-7x-6\)

\(=x^3-x-6x-6\)

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

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

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

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

b) \(2x^3-x^2+5x+3\)

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

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

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

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

c) \(2x^3-5x^2+5x+1\)

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

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

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

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

d) \(2x^3+3x^2+3x+1\)

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

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

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

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

e) \(3x^3-2x^2+5x+2\)

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

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

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

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

f) \(27x^3-27x^2+18x-4\)

\(=27x^3-9x^2-18x^2+6x+12x-4\)

\(=\left(27x^3-9x^2\right)-\left(18x^2-6x\right)+\left(12x-4\right)\)

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

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

2. a. \(A=2x^2-8x-10=2\left(x^2-4x+4\right)-18\)

\(=2\left(x-2\right)^2-18\)

Vì \(\left(x-2\right)^2\ge0\forall x\)\(\Rightarrow2\left(x-2\right)^2-18\ge-18\)

Dấu "=" xảy ra \(\Leftrightarrow2\left(x-2\right)^2=0\Leftrightarrow x-2=0\Leftrightarrow x=2\)

Vậy minA = - 18 <=> x = 2

b. \(B=9x-3x^2=-3\left(x^2-3x+\frac{9}{4}\right)+\frac{27}{4}\)

\(=-3\left(x-\frac{3}{2}\right)^2+\frac{27}{4}\)

Vì \(\left(x-\frac{3}{2}\right)^2\ge0\forall x\)\(\Rightarrow-3\left(x-\frac{3}{2}\right)^2+\frac{27}{4}\le\frac{27}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow-3\left(x-\frac{3}{2}\right)^2=0\Leftrightarrow x-\frac{3}{2}=0\Leftrightarrow x=\frac{3}{2}\)

Vậy maxB = 27/4 <=> x = 3/2

Sửa đề:x³-3x²-4x+12

a,x³-3x²-4x+12

=(x³-3x²)-(4x+12)

=x²(x-3)-4(x-3)

=(x²-4)(x-3)

b,x⁴- 5x² +4

x⁴-4x²-x²+4

(x⁴-x²)-(4x²+4)

x²(x²-1)-4(x²-1)

(x²-4)(x²-1)

Bài 1.

a) x³ - 3x² - 4x + 12 ( mạn phép sửa 13 thành 12, chứ để 13 là không phân tích được :> )

= x²( x - 3 ) - 4( x - 3 )

= ( x - 3 )( x² - 4 )

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

b) x⁴ - 5x² + 4

Đặt t = x²

Đa thức <=> t² - 5t + 4

= t² - t - 4t + 4

= t( t - 1 ) - 4( t - 1 )

= ( t - 1 )( t - 4 )

= ( x² - 1 )( x² - 4 )

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

c) ( x + y + z )³ - x³ - y³ - z³

= ( x + y + z )³ - ( x³ + y³ + z³ )

= ( x + y + z )³ - [ ( x + y + z )³ - 3( x + y )( y + z )( z + x ) ] ( chỗ này bạn xem HĐT tổng ba lập phương nhé )

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

= 3( x + y )( y + z )( z + x )

d) 45 + x³ - 5x² - 9x 

= ( x³ - 5x² ) - ( 9x - 45 )

= x²( x - 5 ) - 9( x - 5 )

= ( x - 5 )( x² - 9 )

= ( x - 5 )( x - 3 )( x + 3 )

e) x⁴ - 2x³ + 3x² - 2x - 3 ( sửa -3x³ -> 3x² )

= x⁴ - x³ - x³ + 3x² - x² + x² - 3x + x - 3

= ( x⁴ - x³ + 3x² ) - ( x³ - x² + 3x ) - ( x² - x + 3 )

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

= ( x² - x - 1 )( x² - x + 3 )

Bài 2.

A = 2x² - 8x - 10

= 2( x² - 4x + 4 ) - 18

= 2( x - 2 )² - 18 

2( x - 2 )² ≥ 0 ∀ x => 2( x - 2 )² - 18 ≥ -18

Đẳng thức xảy ra <=> x - 2 = 0 => x = 2

=> MinA = -18 <=> x = 2

B = 9x - 3x²

= -3( x² - 3x + 9/4 ) + 27/4

= -3( x - 3/2 )² + 27/4

-3( x - 3/2 )² ≤ 0 ∀ x => -3( x - 3/2 )² + 27/4 ≤ 27/4

Đẳng thức xảy ra <=> x - 3/2 = 0 => x = 3/2

=> MaxB = 27/4 <=> x = 3/2

a) Xem lại đề

b) x³ - 4x²y + 4xy² - 9x

= x(x² - 4xy + 4y² - 9)

= x[(x² - 4xy + 4y² - 3²]

= x[(x - 2y)² - 3²]

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

c) x³ - y³ + x - y

= (x³ - y³) + (x - y)

= (x - y)(x² + xy + y²) + (x - y)

= (x - y)(x² + xy + y² + 1)

d) 4x² - 4xy + 2x - y + y²

= (4x² - 4xy + y²) + (2x - y)

= (2x - y)² + (2x - y)

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

e) 9x² - 3x + 2y - 4y²

= (9x² - 4y²) - (3x - 2y)

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

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

f) 3x² - 6xy + 3y² - 5x + 5y

= (3x² - 6xy + 3y²) - (5x - 5y)

= 3(x² - 2xy + y²) - 5(x - y)

= 3(x - y)² - 5(x - y)

= (x - y)[(3(x - y) - 5]

= (x - y)(3x - 3y - 5)