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Решите задачу:

$\frac{1}{ \sqrt{18} } + \frac{1}{ \sqrt{2} } = \frac{\sqrt{2}+\sqrt{18}}{ \sqrt{18}\sqrt{2} } = \frac{\sqrt{2}+\sqrt{18}}{\sqrt{36}} = \frac{\sqrt{2}+\sqrt{18}}{6}= \frac{\sqrt{2}+\sqrt{9} \sqrt{2} }{6} = \\ \\ = \frac{ \sqrt{2}(1+ \sqrt{9}) }{6}= \frac{ \sqrt{2}(1+3) }{6} = \frac{4 \sqrt{2} }{6} = \frac{2 \sqrt{2} }{3}$