Question

doar A va rog *CALCULATI SUMA INVERSELOR NUMERELOR a,b,c unde:​

Answer 1

Răspuns:

$\frac{1}{\sqrt{8} }+\frac{1}{\sqrt{18}}+\frac{1}{\sqrt{72} }=\frac{1}{\sqrt{4*2} }+\frac{1}{\sqrt{9*2}}+\frac{1}{\sqrt{36*2} }=\frac{1}{2*\sqrt{2} }+\frac{1}{3*\sqrt{2}}+\frac{1}{6*\sqrt{2} }=\frac{\sqrt{2} }{4 }+\frac{\sqrt{2} }{6}+\frac{\sqrt{2} }{12}=\frac{3*\sqrt{2} }{12 }+\frac{2*\sqrt{2} }{12}+\frac{\sqrt{2} }{12}= \frac{6*\sqrt{2} }{12}=\frac{\sqrt{2} }{2}$

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