Chịu

1: Sửa đề: 3x-5

\(=\dfrac{-x^2\left(3x-5\right)-3\left(3x-5\right)}{3x-5}=-x^2-3\)

2: \(=\dfrac{5x^4-5x^3+14x^3-14x^2+12x^2-12x+8x-8}{x-1}\)

=5x^2+14x^2+12x+8

3: \(=\dfrac{5x^3+10x^2+4x^2+8x+4x+8}{x+2}=5x^2+4x+4\)

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

5: \(=\dfrac{x^2\left(5-3x\right)+3\left(5-3x\right)}{5-3x}=x^2+3\)

Chia nhỏ ra ik ạ

\(\left(2x+3\right)\left(2x-3\right)-4x\left(x+5\right)=4x^2-9-4x^2-20x=-20x-9\)

\(5x\left(x-3\right)+\left(x-2\right)^2=5x^2-15x+x^2-4x+4=6x^2-19x+4\)

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

a) 6x³ + 3x² + 4x + 2

= ( 6x³ + 3x² ) + ( 4x + 2 )

= 3x²( 2x + 1 ) + 2( 2x + 1 )

= ( 2x + 1 )( 3x² + 2 )

=> (  6x³ + 3x² + 4x + 2 ) : ( 3x² + 2 ) = 2x + 1

b) 2x³ - 26x - 24

= 2( x³ - 13x - 12 )

= 2( x³ + 4x² - 4x² + 3x - 16x - 12 )

= 2[ ( x³ + 4x² + 3x ) - ( 4x² + 16x + 12 ) ]

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

= 2( x² + 4x + 3 )( x - 4 )

=> ( 2x³ - 26x - 24 ) : ( x² + 4x + 3 ) = 2( x - 4 ) = 2x - 8

c) x³ - 7x + 6 

= x³ - 3x² + 3x² + 2x - 9x - 6

= ( x³ - 3x² + 2x ) + ( 3x² - 9x + 6 )

= x( x² - 3x + 2 ) + 3( x² - 3x + 2 )

= ( x² - 3x + 2 )( x + 3 )

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

a,\(\left(6x^3+3x^2+4x+2\right)\div\left(3x^2+2\right)\)

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

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

\(=2x+1\)