Đáp án:
$a. A = \sqrt[]{\frac{1}{2}} + \sqrt[]{2} - 3\sqrt[]{4,5} + \sqrt[]{12,5} + \sqrt[]{18}$
$⇔ A = \frac{\sqrt[]{2}}{2} + \sqrt[]{2} - 3\sqrt[]{\frac{9}{2}} + \sqrt[]{\frac{25}{2}} + \sqrt[]{18}$
$⇔ A = \frac{3\sqrt[]{2}}{2} - 3.\frac{3}{\sqrt[]{2}} + \frac{5}{\sqrt[]{2}} + 3\sqrt[]{2}$
$⇔ A = \frac{3\sqrt[]{2}}{2} - \frac{9\sqrt[]{2}}{2} + \frac{5\sqrt[]{2}}{2} + 3\sqrt[]{2}$
$⇔ A = \frac{5\sqrt[]{2}}{2}$
$b. B = ab\sqrt[]{4ab} + \sqrt[]{9a^{3}b^{3}} - 3a\sqrt[]{25ab^{3}}$ $( a , b > 0 )$
$⇔ B = 2ab\sqrt[]{ab} + 3ab\sqrt[]{ab} - 15ab\sqrt{ab}$
$⇔ B = - 10ab\sqrt[]{ab}$
( do $a , b > 0 ⇒ \sqrt[]{a^{2}} = a , \sqrt[]{b^{2}} = b , \sqrt[]{a^{2}b^{2}} = ab$ )
$c. C = \sqrt[]{a-\sqrt[]{4a}+1} + \sqrt[]{a+\sqrt[]{4a}+1}$ $( 0 < a < 1 )$
$⇔ C = \sqrt[]{a-2\sqrt[]{a}+1} + \sqrt[]{a+2\sqrt[]{a}+1}$
$⇔ C = \sqrt[]{(\sqrt[]{a}-1)^{2}} + \sqrt[]{(\sqrt[]{a}+1)^{2}}$
$⇔ C = | \sqrt[]{a} - 1 | + | \sqrt[]{a} + 1 |$
$⇔ C = 1 - \sqrt[]{a} + \sqrt[]{a} + 1$
$⇔ C = 2$
( do $0 < a < 1 ⇒ \sqrt[]{a} - 1 < 0 , \sqrt[]{a} + 1 > 0 ⇒ | \sqrt[]{a} - 1 | = 1 - \sqrt[]{a} , | \sqrt[]{a} + 1 | = \sqrt[]{a} + 1$ )
