a: \(=2x^3:\dfrac{-3}{2}x+4x:\dfrac{3}{2}x-5:\dfrac{3}{2}\)

=-4/3x^2+8/3-10/3

=-4/3x^2-2/3

d: \(\dfrac{3x^3-5x+2}{x-3}=\dfrac{3x^3-9x^2+9x^2-27x+22x-66+68}{x-3}\)

\(=3x^2+9x+22+\dfrac{68}{x-3}\)

a) –4x(5x^2 – 2xy + y^2)

= -20x^3 + 8x^2y - 4xy^2

b) (4x – 1)(2x^2 – x – 1)

= 8x^3 - 4x^2 - 4x - 2x^2 + x + 1

= 8x^3 - 6x^2 - 3x + 1

a) ( 3x - 2 )( 4x + 5 ) - 6x( 2x - 1 )

= 12x² + 7x - 10 - 12x² + 6x

= 13x - 10

b) ( 2x - 5 )² - 4( x + 3 )( x - 3 )

= 4x² - 20x + 25 - 4( x² - 9 )

= 4x² - 20x + 25 - 4x² + 36

= 61 - 20x

c) 2x³ - 5x² + 7x - 6

= 2x³ - 3x² - 2x² + 3x + 4x - 6

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

= ( 2x - 3 )( x² - x + 2 )

=> ( 2x³ - 5x² + 7x - 6 ) : ( 2x - 3 ) = x² - x + 2

a, (3x - 2 ) (4x + 5) - 6x (2x -1) = ( 7x + 15x -8x - 10 ) - ( 12x² -6x ) = 7x² + 15x - 8x -10 -12x² + 6x = -5x² + x - 10 

Mn ơi làm hộ mk vs mai mk kiểm tra

Bài 3:

A(x)⋮B(x)

=>\(3x^2+5x+m\) ⋮x-2

=>\(3x^2-6x+11x-22+m+22\) ⋮x-2

=>m+22=0

=>m=-22

Bài 2:

a: \(2x^3-8x^2+8x\)

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

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

b: 2xy+2x+yz+z

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

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

c: \(x^2+2x+1-y^2\)

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

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

Câu 1:

a:\(\left(4x-1\right)\left(2x^2-x-1\right)\)

\(=8x^3-4x^2-4x-2x^2+x+1\)

\(=8x^3-6x^2-3x+1\)

b: \(\left(4x^3+8x^2-2x\right):2x\)

\(=\frac{4x^3}{2x}+\frac{8x^2}{2x}-\frac{2x}{2x}\)

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

c: \(\left(6x^3-7x^2-16x+12\right):\left(2x+3\right)\)

\(=\left(6x^3+9x^2-16x^2-24x+8x+12\right):\left(2x+3\right)\)

\(=\left\lbrack3x^2\left(2x+3\right)-8x\left(2x+3\right)+4\left(2x+3\right)\right\rbrack:\left(2x+3\right)\)

\(=3x^2-8x+4\)

a, \(5x^2-10xy+5y^2=5\left(x^2-2xy+y^2\right)=5.\left(x-y\right)^2\)

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

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

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

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