Điền vào chỗ trống: $\dfrac{{2x - 6}}{{x + 3}} - .... = \dfrac{{x + 1}}{2}$.

$3x(x - 2)$    

$x - 2$

$3{x^2}(x - 2)$

$3x{(x - 2)^2}$

$\dfrac{x}{{x + 1}} - \dfrac{2}{{x + 1}}$.

$\dfrac{{2x}}{{x + 1}} - \dfrac{2}{{x + 1}}$.

$\dfrac{x}{{x - 1}} + \dfrac{1}{{x - 1}}$.

$\dfrac{x}{{x + 1}} + \dfrac{1}{{ - x - 1}}$.

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

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

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

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

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

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

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

$\dfrac{{22{x^3}{y^3}}}{5}$.

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

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

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

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

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

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

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

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