Hướng dẫn trả lời:
$\text{a) $\sqrt{9 - \sqrt{17}}$ - $\sqrt{9 + \sqrt{17}}$.}$
$\text{Đặt N = $\sqrt{9 - \sqrt{17}}$ - $\sqrt{9 + \sqrt{17}}$.}$
$\text{⇒ $N^2$ = $(\sqrt{9 - \sqrt{17}}$ - $\sqrt{9 + \sqrt{17}})^2$.}$
$\text{$N^2$ = $9 - \sqrt{17} - 2.\sqrt{9 - \sqrt{17}}.\sqrt{9 + \sqrt{17}} + 9 + \sqrt{17}$.}$
$\text{$N^2$ = $18 - 2.\sqrt{(9 - \sqrt{17}).(9 + \sqrt{17})}$.}$
$\text{$N^2$ = $18 - 2.\sqrt{81 - 17}$.}$
$\text{$N^2$ = $18 - 2.\sqrt{64}$.}$
$\text{$N^2$ = $18 - 2.8$.}$
$\text{$N^2$ = $18 - 16$.}$
$\text{$N^2$ = 2.}$
$\text{→ N = ± $\sqrt{2}$.}$
$\text{Mà N < 0 (Vì $\sqrt{9 - \sqrt{17}}$ < $\sqrt{9 + \sqrt{17}}$).}$
Nên N = - $\sqrt{2}$.
$\text{Vậy giá trị của $\sqrt{9 - \sqrt{17}}$ - $\sqrt{9 + \sqrt{17}}$ là}$ - $\sqrt{2}$.
$\text{b) $\sqrt{11 - 2\sqrt{30}}$ + $\sqrt{13 + 2\sqrt{42}}$.}$
Ta có: $\sqrt{11 - 2\sqrt{30}}$ = $\sqrt{6 - 2\sqrt{6}.\sqrt{5} + 5}$ = $\sqrt{(\sqrt{6}- \sqrt{5})^2}$ = $|\sqrt{6} - \sqrt{5}|$ = $\sqrt{6} - \sqrt{5}$. (Vì $\sqrt{6} - \sqrt{5}$ > 0)
Và: $\sqrt{13 + 2\sqrt{42}}$ = $\sqrt{7 + 2\sqrt{7}.\sqrt{6} + 6}$ = $\sqrt{(\sqrt{7}+ \sqrt{6})^2}$ = $|\sqrt{7} + \sqrt{6}|$ = $\sqrt{7} + \sqrt{6}$. (Vì $\sqrt{7} + \sqrt{6}$ > 0)
$\text{Nên $\sqrt{11 - 2\sqrt{30}}$ + $\sqrt{13 + 2\sqrt{42}}$.}$
$\text{= ($\sqrt{6} - \sqrt{5}$) + ($\sqrt{7} + \sqrt{6}$).}$
$\text{= $\sqrt{6} - \sqrt{5}$ + $\sqrt{7} + \sqrt{6}$.}$
$\text{= $2\sqrt{6} - \sqrt{5} + \sqrt{7}$.}$
