$A = \frac{2a²+4}{1-a³} - \frac{1}{1+\sqrt{a}} - \frac{1}{1-\sqrt{a}}$
$= \frac{2a²+4}{(1+\sqrt{a})(1-\sqrt{a})(1 + a + a²)}- \frac{1}{1+\sqrt{a}} - \frac{1}{1-\sqrt{a}}$
$= \frac{2a²+4}{(1+\sqrt{a})(1-\sqrt{a})(1 + a + a²)} - \frac{(1-\sqrt{a})(1 + a + a²)}{(1+\sqrt{a})(1-\sqrt{a})(1 + a + a²)} - \frac{(1+\sqrt{a})(1 + a + a²)}{(1+\sqrt{a})(1-\sqrt{a})(1 + a + a²)}$
$= \frac{2a²+4-(1-\sqrt{a})(1 + a + a²) - (1+\sqrt{a})(1 + a + a²) }{(1+\sqrt{a})(1-\sqrt{a})(1 + a + a²)}$
$ = \frac{2a²+4-(1 + a + a²)(1 - \sqrt{a} +1+\sqrt{a}) }{(1+\sqrt{a})(1-\sqrt{a})(1 + a + a²)}$
$ = \frac{2a²+4-2(1 + a + a²)}{(1+\sqrt{a})(1-\sqrt{a})(1 + a + a²)}$
$ = \frac{2a²+4 - 2 - 2a - 2a²}{(1-a)(1 + a + a²)}$
$ = \frac{2-2a}{(1-a)(1 + a + a²)}$
$ = \frac{2(1-a)}{(1-a)(1 + a + a²)} = \frac{2}{1 + a + a²}$
b)
Ta có: $a² + a + 1 = a² + 2.\frac{1}{2}.a + \frac{1}{4} + \frac{3}{4} = (a + \frac{1}{2})² + \frac{3}{4}$
Theo điều kiện: $a ≥ 0$
$⇔ a + \frac{1}{2} ≥ \frac{1}{2}$
$⇔ (a + \frac{1}{2})² ≥ \frac{1}{4}$
$⇔ (a + \frac{1}{2})² + \frac{3}{4} ≥ 1$
$⇔ a² + a + 1 ≥ 1$
$⇔ \frac{1}{a² + a + 1} ≤ 1$
$⇔ \frac{2}{1 + a + a²} ≤ 2$
Dấu "=" xảy ra khi: $a = 0$
Vậy GTLN đạt được là 2 khi $a = 0$
