(i) \((\frac{1}{2})^{-1}+(\frac{1}{3})^{-1}+(\frac{1}{4})^{-1}\)

⇒ (2)+(3)+(4)[Using \(a^{-n}=\frac{1}{a^{n}}\)]

⇒ 2+3+4 = 9

(ii) \((\frac{1}{2})^{-2}+(\frac{1}{3})^{-2}+(\frac{1}{4})^{-2}\)

⇒ \((2^{2})+(3^{2})+(4^{2})\)[Using \(a^{-n}=\frac{1}{a^{n}}\);\(a^{2}=a\times a\)]

⇒ 4+9+16 = 29

(iii) \((2^{-1}\times 4^{-1})÷2^{2}\)

⇒ \((\frac{1}{2}\times \frac{1}{4})÷(\frac{1}{2^{2}})\)[Using \(a^{-n}=\frac{1}{a^{n}}\)and \(\frac{1}{a}\div \frac{1}{b}\) = \(\frac{1}{a}\times \frac{b}{1}\)]

⇒ \((\frac{1\times 1}{2\times 4})\times 2^{2}\)[Using \(\frac{1}{a}\times \frac{b}{1}\)= \(\frac{1}{ab}\);\(a^{2}=a\times a\)] 

⇒ \(\frac{1}{8}\times 4\) = \(\frac{1}{2}\)

(iv) \((5^{-1}\times 2^{-1})÷6^{-1}\)

⇒ \((\frac{1}{5}\times \frac{1}{2})÷(\frac{1}{6})\)[Using  \(a^{-n}=\frac{1}{a^{n}}\)and \(\frac{1}{a}\div \frac{1}{b}\) = \(\frac{1}{a}\times \frac{b}{1}\)]

⇒ \((\frac{1\times 1}{5\times 2})\times (\frac{6}{1})\)[Using  \(\frac{1}{a}\times \frac{b}{1}\)= \(\frac{1}{ab}\)] 

⇒   \(\frac{1}{10}\times 6\) = \(\frac{6}{10}\) = \(\frac{3}{5}\)