a) Đặt $A= \sqrt{6+\sqrt{6+\sqrt{6+\dots}}}$
$\Rightarrow A^2 = 6 + \sqrt{6+\sqrt{6+\sqrt{6+\dots}}}$
$\Leftrightarrow A^2 = 6 + A$
$\Leftrightarrow (A + 2)(A-3)=0$
$\Leftrightarrow A = 3\qquad (Do \,\,A>0)$
b) $x = \sqrt{\dfrac{3}{5}} + \sqrt{\dfrac{3}{5}} = \dfrac{8}{\sqrt{15}}$
$\Rightarrow a\sqrt{15} = 8$
Ta được:
$A = \sqrt{15a^2 - 8a\sqrt{15} + 16}$
$\to A = \sqrt{(a\sqrt{15} - 4)^2}$
$\to A = |a\sqrt{15} - 4|$
$\to A = |8 - 4| = 4$