Đa thức thích hợp để điền vào chỗ trống trong đẳng thức \(\dfrac{{{x^3} - 8}}{{......}} = \dfrac{{{x^2} + 2x + 4}}{{3x}}\) là:

\(\dfrac{x}{{x + 1}} - \dfrac{2}{{x + 1}}\).

\(\dfrac{{2x}}{{x + 1}} - \dfrac{2}{{x + 1}}\).

\(\dfrac{x}{{x - 1}} + \dfrac{1}{{x - 1}}\).

\(\dfrac{x}{{x + 1}} + \dfrac{1}{{ - x - 1}}\).

\(4{\left( {x + 3} \right)^2}\)

\(4\left( {x - 3} \right)\left( {x + 3} \right)\)

\(\left( {x - 3} \right)\left( {x + 3} \right)\)

\(4{\left( {x - 3} \right)^2}\)

\(\dfrac{{11{x^2}{y^3}}}{5}\).

\(\dfrac{{22{x^2}{y^3}}}{5}\).

\(\dfrac{{22{x^2}{y^3}}}{{25}}\).

\(\dfrac{{22{x^3}{y^3}}}{5}\).

$\dfrac{{ - {x^2} + 15}}{{2(x + 3)}}$

$\dfrac{{{x^2} - 15}}{{2(x + 3)}}$

$\dfrac{{ - {x^2} - 15}}{{2(x + 3)}}$

Cả A, B, C đều sai

$\dfrac{{2x - 9}}{{{x^2} - 9}}$

$\dfrac{{2x - 3}}{{{x^2} - 9}}$

$\dfrac{{2x - 9}}{{x - 3}}$

$\dfrac{{x - 6}}{{{x^2} - 9}}$

\({\left( {x + 3} \right)^3}\)

\(3{\left( {x + 3} \right)^2}\)

\(3{\left( {x + 3} \right)^3}\)

\({\left( {x + 3} \right)^4}\)