Question

Answer 1

`#3107.101107`

a)

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

`<=>`\(\dfrac{2x^2\left(4x\right)}{-2x^2}=\dfrac{-x^3\left(4x\right)}{x^3}\)

`<=>`\(-4x=-4x\)

`<=> 4x = 4x` (luôn đúng)

Vậy, `(8x^3)/(-2x^2) = (-4x^4)/(x^3)`

b)

\(\dfrac{x-2}{2x-4}=\dfrac{x^2-4}{2x^2-8}\)

`<=> (x - 2)/( 2(x - 2)) = (x^2 - 4)/( 2(x^2 - 4))`

`<=> 1/2 = 1/2` (luôn đúng)

Vậy, `(x - 2)/(2x - 4) = (x^2 - 4)/(2x^2 - 8).`