Sederhanakan lalu rasionalkan bentuk akar berikut ini. a.

a. (\(\sqrt{6}+\sqrt{2}\))/2, b. (\(\sqrt{14}+\sqrt{10}\))/2, c. (3\(\sqrt{2}\)-\(\sqrt{6}\))/2

Pembahasan

Untuk menyederhanakan dan merasionalkan bentuk akar:
a. \(\sqrt{2 + \sqrt{3}}\)
Kita cari dua bilangan a dan b sehingga (\(\sqrt{a} + \sqrt{b}\))^2 = a + b + 2\(\sqrt{ab}\) = 2 + \(\sqrt{3}\).
Ini sulit disederhanakan secara langsung. Namun, kita bisa menggunakan identitas \(\sqrt{A \pm \sqrt{B}} = \sqrt{\frac{A+C}{2}} \pm \sqrt{\frac{A-C}{2}}\), di mana C = \(\sqrt{A^2 - B}\).
Untuk \(\sqrt{2 + \sqrt{3}}\), A=2, B=3. C = \(\sqrt{2^2 - 3} = \sqrt{4-3} = \sqrt{1} = 1\).
Maka, \(\sqrt{2 + \sqrt{3}} = \sqrt{\frac{2+1}{2}} + \sqrt{\frac{2-1}{2}} = \sqrt{\frac{3}{2}} + \sqrt{\frac{1}{2}} = \frac{\sqrt{3}}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{\sqrt{3}+1}{\sqrt{2}} = \frac{(\sqrt{3}+1)\sqrt{2}}{2} = \frac{\sqrt{6}+\sqrt{2}}{2}\).
b. \(\sqrt{6 + \sqrt{35}}\)
A=6, B=35. C = \(\sqrt{6^2 - 35} = \sqrt{36-35} = \sqrt{1} = 1\).
Maka, \(\sqrt{6 + \sqrt{35}} = \sqrt{\frac{6+1}{2}} + \sqrt{\frac{6-1}{2}} = \sqrt{\frac{7}{2}} + \sqrt{\frac{5}{2}} = \frac{\sqrt{7}}{\sqrt{2}} + \frac{\sqrt{5}}{\sqrt{2}} = \frac{\sqrt{7}+\sqrt{5}}{\sqrt{2}} = \frac{(\sqrt{7}+\sqrt{5})\sqrt{2}}{2} = \frac{\sqrt{14}+\sqrt{10}}{2}\).
c. \(\sqrt{6 - \sqrt{27}}\)
A=6, B=27. C = \(\sqrt{6^2 - 27} = \sqrt{36-27} = \sqrt{9} = 3\).
Maka, \(\sqrt{6 - \sqrt{27}} = \sqrt{\frac{6+3}{2}} - \sqrt{\frac{6-3}{2}} = \sqrt{\frac{9}{2}} - \sqrt{\frac{3}{2}} = \frac{\sqrt{9}}{\sqrt{2}} - \frac{\sqrt{3}}{\sqrt{2}} = \frac{3}{\sqrt{2}} - \frac{\sqrt{3}}{\sqrt{2}} = \frac{3-\sqrt{3}}{\sqrt{2}} = \frac{(3-\sqrt{3})\sqrt{2}}{2} = \frac{3\sqrt{2}-\sqrt{6}}{2}\).