\(=x^3+1\)

\(=x^3+1\)

(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 - 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

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

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

\(=\dfrac{1}{x+1}\)

(1-x)^2-x(x-1)

Đặt \(x^2+1=a\)

Ta có: \(\dfrac{1}{x^2-x+1}-\dfrac{x^2+2}{x^2+1}+1\)

\(=\dfrac{1}{a-x}+\dfrac{a+1}{a}+1\)

\(=\dfrac{a}{a\left(a-x\right)}+\dfrac{\left(a+1\right)\left(a-x\right)}{a\left(a-x\right)}+\dfrac{a\left(a-x\right)}{a\left(a-x\right)}\)

\(=\dfrac{a+a^2-ax+a-x+a^2-ax}{a\left(a-x\right)}\)

\(=\dfrac{2a^2+2a-2ax-x}{a\left(a-x\right)}\)

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

\(=\dfrac{2\left(x^4+2x^2+1\right)+2x^2+2-2x^3-2x-x}{\left(x^2+1\right)\left(x^2-x+1\right)}\)

\(=\dfrac{2x^4+4x^2+2+2x^2+2-2x^3-3x}{\left(x^2+1\right)\left(x^2-x+1\right)}\)

\(=\dfrac{2x^4-2x^3+6x^2-3x+4}{\left(x^2+1\right)\left(x^2-x+1\right)}\)

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