a = √6+√6+√6+... b = √20+√20+√20+... maka, a + b adalah

3
[tex]3 \sqrt{6} + 3 \sqrt{20} = 3 \sqrt{6} + 3 \sqrt{4 \times 5} = 3 \sqrt{6} + 6 \sqrt{5} = 3 (\sqrt{6} + 2 \sqrt{5} [/tex]

[tex] a = \sqrt{6 + \sqrt{6 + \sqrt{6 + ...}}} \\ a^2 = 6 + \sqrt{6 + \sqrt{6 + \sqrt{6 + ...}}} \\ a^2 = 6 + a \\ a^2 - a - 6 = 0 \\ (a - 3)(a + 2) = 0 \\ a = 3 \\ {atau} \\ a = -2 [/tex]
Namun, hasil dari akar tidak pernah negatif, sehingga a = 3

[tex] b = \sqrt{20 + \sqrt{20 + \sqrt{20 + ...}}} \\ b^2 = 20 + \sqrt{20 + \sqrt{20 + \sqrt{20 + ...}}} \\ b^2 = 20 + b \\ b^2 - b - 20 = 0 \\ (b - 5)(b + 4) = 0 [/tex]
b = 5


Jadi, nilai dari
a + b
= 3 + 5
= 8