\(=x^3+1\)

\(=x^3+1\)

(x – 1)(x + 1)(x + 2)

= ( x 2  + x – x – 1)(x + 2)

= ( x 2 – 1)(x + 2)

=  x 2 ( x + 2) – 1.(x +2)

=  x 3  + 2 x 2  – x – 2

`@` `\text {Ans}`

`\downarrow`

`x(1-x) + (x-1)^2`

`= x-x^2 + x^2 - 2x + 1`

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

`= -x+1`

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

\(=x^2-4x+4+3x-x^2+3-x=-2x+7\)

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

\(\left(x+2\right)^2-\left(x-3\right)\left(x+1\right)=x^2+4x+4-\left(x^2+x-3x-3\right)=x^2+4x+4-x^2-x+3x+3=6x+7.\)

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

(x - 1/2 )(x + 1/2 )(4x - 1)

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

= ( x 2  - 1/4 )(4x - 1)

= 4 x 3  –  x 2  – x + 1/4

Bài 1: 

b: \(=\dfrac{x+3-4-x}{x-2}=\dfrac{-1}{x-2}\)

Bài 2: 

a: \(=\dfrac{x+1}{2\left(x+3\right)}+\dfrac{2x+3}{x\left(x+3\right)}\)

\(=\dfrac{x^2+x+4x+6}{2x\left(x+3\right)}=\dfrac{x^2+5x+6}{2x\left(x+3\right)}=\dfrac{x+2}{2x}\)

d: \(=\dfrac{3}{2x^2y}+\dfrac{5}{xy^2}+\dfrac{x}{y^3}\)

\(=\dfrac{3y^2+10xy+2x^3}{2x^2y^3}\)

e: \(=\dfrac{x^2+2xy+x^2-2xy-4xy}{\left(x+2y\right)\left(x-2y\right)}=\dfrac{2x^2-4xy}{\left(x+2y\right)\cdot\left(x-2y\right)}=\dfrac{2x}{x+2y}\)