Bài 3: 

\(f\left(x\right)=9x^3-\frac{1}{3}x+3x^2-3x+\frac{1}{3}x^2-\frac{1}{9}x^3-3x^2-9x+27+3x\) 

\(f\left(x\right)=\left(9x^3-\frac{1}{9}x^3\right)-\left(\frac{1}{3}x+3x+9x-3x\right)+\left(3x^2-3x^2\right)+27\) 

\(f\left(x\right)=\frac{80}{9}x^3-\frac{28}{3}x+27\) 

Thay x = 3 vào đa thức, ta có:

\(f\left(3\right)=\frac{80}{9}.3^3-\frac{28}{3}.3+27\) 

\(f\left(3\right)=240-28+27=239\)

Vậy đa thức trên bằng 239 tại x = 3

Thay x = -3 vào đa thức. ta có:

\(f\left(-3\right)=\frac{80}{9}.\left(-3\right)^3-\frac{28}{3}.\left(-3\right)+27\)

\(f\left(-3\right)=-240+28+27=-185\)

Bài 4: \(f\left(x\right)=2x^6+3x^2+5x^3-2x^2+4x^4-x^3+1-4x^3-x^4\)

\(f\left(x\right)=2x^6+\left(3x^2-2x^2\right)+\left(5x^3-x^3-4x^3\right)+\left(4x^4-x^4\right)\)

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

Thay x=1 vào đa thức, ta có:

\(f\left(1\right)=2.1^6+1^2+3.1^4=2+1+3=6\)

Đa thức trên bằng 6 tại x =1

Thay x = - 1 vào đa thức, ta có:

\(f\left(-1\right)=2.\left(-1\right)^6+\left(-1\right)^2+3.\left(-1\right)^4=2+1+3=6\)

Đa thức trên có nghiệm = 0

a/ 2x\(^{^{ }3}\)-3\(^{^{ }3}\)-2x\(^3\)-1\(^{^{ }3}\)=-28

b/x\(^{^{ }3}\)+2\(^{^{ }3}\)-x\(^3\)+2=10

c/3x\(^3\)+5\(^3\)-3x(3x\(^2\)-1)=3x\(^3\)+5\(^3\)-3x\(^3\)+3x=125+3x

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

Bài 2

a) 4x(x-3)-3x+9

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

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

b) x³+2x²-2x-4

=(x³+2x²)-(2x+4)

=x²(x+2)-2(x+2)

=(x+2)(x²-2)

c) 4x²-4y+4y-1

=4x²-1

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

d) x⁵-x

=x(x⁴-1)

=x(x²-1)(x²+1)

a) 4x(x-3)-3x+9

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

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

b)x³ + 2x² - 2x - 4

= x²(x + 2) - 2(x + 2)

= (x+2)(x²-2)

c) 4x² - 4y +4y -1

= [(2x)²-1²] + (-4y+4y)

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

d) x⁵-x

= x(x⁴ - 1)

\(a,-5x\left(x-3\right)\left(2x+4\right)-\left(x+3\right)\left(x-3\right)+\left(5x-2\right)\left(3x+4\right)\)

\(=-5x\left(2x^2-x-12\right)-\left(x^2-9\right)+15x^2+20x-6x-8\)

\(=-10x^3+5x^2+60x-x^2+9+15x^2+20x-6x-8\)

\(=-10x^3+19x^2+74x+1\)

\(b,\left(4x-1\right)x\left(3x+1\right)-5x^2.x\left(x-3\right)-\left(x-4\right)x\left(x-5\right)\)\(-7\left(x^3-2x^2+x-1\right)\)

\(=\left(4x^2-x\right)\left(3x+1\right)-5x^4-15x^3-\left(x^2-4x\right)\left(x-5\right)\)\(-7x^3+14x^2-7x+7\)

\(=12x^3+x^2-x-5x^4-15x^3-x^3+9x^2+20x\)\(-7x^3+14x^2-7x+7\)

\(=-5x^4-11x^3+24x^2+12x+7\)

\(c,\left(5x-7\right)\left(x-9\right)-\left(3-x\right)\left(2-5x\right)-2x\left(x-4\right)\)

\(=5x^2-52x+63-6+17x-5x^2-2x^2+8x\)

\(=-2x^2-27x+57\)

\(d,\left(5x-4\right)\left(x+5\right)-\left(x+1\right)\left(x^2-6\right)-5x+19\)

\(=5x^2+21x-20-x^3-x^2+6x+6-5x+19\)

\(=-x^3+4x^2+22x+5\)

\(e,\left(9x^2-5\right)\left(x-3\right)-3x^2\left(3x+9\right)-\left(x-5\right)\left(x+4\right)-9x^3\)

\(=9x^3-27x^2-5x+15-9x^3-27x^2-x^2+x+20-9x^3\)

\(=-9x^3-55x^2+4x+35\)

\(g,\left(x-1\right)^2-\left(x+2\right)^2\)

\(=x^2-2x+1-x^2-4x-4\)

\(=-6x-3\)

a, =(2x+3-x-2).(2x+3+x+2)/x².(x+1)=(x+1).(3x+5)/x².(x+1)=3x+5/x²

^b,  =4x².(x-2)+3.(x-2)/4x².(3x+1)+3.(3x+1)=(4x²+3).(x-2)/(4x²+3).(3x+1)=x-2/3x+1

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

A = \(x^2-6x+9-4x^2+1=-3x^2-6x+10\)

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

B = \(4x^2-12x+9-2x^2-x+2x+1\)

B = \(2x^2-11x+10\)

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

C = \(4x\left(x^2-6x+9\right)-16+16x-4x^2\)

C = \(4x^3-24x^2+36x-16+16x-4x^2\)

C = \(4x^3-28x^2+52x-16\)

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

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

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

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

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

a) (x+2)²+x(x-4)

   =x²+4x+4+x²-4x

   =2x²+4

b)(x-3)²-(x+3)(x-4)

  =x²-6x+9-x²+4x-3x+12

  =-5x+12

c) (3x+1)²+3x(2-4x)

   =9x²+6x+1+6x-12x²

   =-3x²+12x+1

d) (2x-4y)²-(2x-3)(2x-3y)

  =4x²-16xy+16y²-4x²+6xy+6x-9y

  =16y²-10xy+6x-9y

Bài 1.

a) -2x( -3x + 2 ) - ( x + 2 )²

= 6x² - 4x - ( x² + 4x + 4 )

= 6x² - 4x - x² - 4x - 4

= 5x² - 8x - 4

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

= x³ + 8 + 2( x + 1 )( x - 1 )

= x³ + 8 + 2( x² - 1 )

= x³ + 8 + 2x² - 2

= x³ + 2x² + 6

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

= 4x² - 4x + 1 - 8x² + 2 + 4x² + 4x + 1

= 4

d) x² - 3x + xy - 3y

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

= ( x - 3 )( x + y )

Bài 2.

a) 4x² - 4xy + y² = ( 2x - y )²

b) 9x³ - 9x²y - 4x + 4y

= 9x²( x - y ) - 4( x - y )

= ( x - y )( 9x² - 4 )

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

c) x³ + 2 + 3( x³ - 2 )

= x³ + 2 + 3x³ - 6

= 4x³ - 4

= 4( x³ - 1 )

= 4( x - 1 )( x² + x + 1 )

Bài 3.

2( x - 2 ) = x² - 4x + 4

⇔ ( x - 2 )² - 2( x - 2 ) = 0

⇔ ( x - 2 )( x - 2 - 2 ) = 0

⇔ ( x - 2 )( x - 4 ) = 0

⇔ x = 2 hoặc x = 4