Giải thích các bước giải:
$\dfrac{2a^2+4}{1-a^3}-\dfrac{1}{1+\sqrt{a}}-\dfrac{1}{1-\sqrt{a}}$
$=\dfrac{2a^2+4}{1-a^3}-(\dfrac{1}{1+\sqrt{a}}+\dfrac{1}{1-\sqrt{a}})$
$=\dfrac{2a^2+4}{1-a^3}-(\dfrac{1+\sqrt{a}+1-\sqrt{a}}{(1+\sqrt{a})(1-\sqrt{a})})$
$=\dfrac{2(a^2+2)}{(1-a)(1+a+a^2)}-\dfrac{2}{1-a}$
$=\dfrac{2}{1-a}(\dfrac{a^2+2}{1+a+a^2}-1)$
$=\dfrac{2}{1-a}.\dfrac{a^2+2-a^2-a-1}{1+a+a^2}$
$=\dfrac{2}{1-a}.\dfrac{1-a}{1+a+a^2}$
$=\dfrac{2}{1+a+a^2}$
