$\frac{2 a^{2}+4}{1-a^3}$- $\frac{1}{1+ √a}$ – $\frac{1}{1- √a}$

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1. Giải thích các bước giải:

   $\dfrac{2a^2+4}{1-a^3}-\dfrac{1}{1+\sqrt{a}}-\dfrac{1}{1-\sqrt{a}}$ 

   $=\dfrac{2a^2+4}{1-a^3}-(\dfrac{1}{1+\sqrt{a}}+\dfrac{1}{1-\sqrt{a}})$ 

   $=\dfrac{2a^2+4}{1-a^3}-(\dfrac{1+\sqrt{a}+1-\sqrt{a}}{(1+\sqrt{a})(1-\sqrt{a})})$ 

   $=\dfrac{2(a^2+2)}{(1-a)(1+a+a^2)}-\dfrac{2}{1-a}$ 

   $=\dfrac{2}{1-a}(\dfrac{a^2+2}{1+a+a^2}-1)$ 

   $=\dfrac{2}{1-a}.\dfrac{a^2+2-a^2-a-1}{1+a+a^2}$ 

   $=\dfrac{2}{1-a}.\dfrac{1-a}{1+a+a^2}$ 

   $=\dfrac{2}{1+a+a^2}$ 

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