Question : $\frac{1}{\sqrt a}-\frac{1}{\sqrt b}=0$, then the value of $\frac{1}{a}+\frac{1}{ b}$ is:
Option 1: $\frac{1}{\sqrt{ab}}$
Option 2: $\sqrt{ab}$
Option 3: $\frac{2}{\sqrt{ab}}$
Option 4: $\frac{1}{2\sqrt{ab}}$
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Correct Answer: $\frac{2}{\sqrt{ab}}$
Solution : Given: $\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{b}}=0$ ⇒ $(\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{b}})^2=0^2$ ⇒ $(\frac{1}{{a}}+\frac{1}{{b}}-\frac{2}{\sqrt{ab}})=0$ $\therefore (\frac{1}{{a}}+\frac{1}{{b}})=\frac{2}{\sqrt{ab}}$ Hence, the correct answer is $\frac{2}{\sqrt{ab}}$.
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Question : If $2x=\sqrt{a}+\frac{1}{\sqrt{a}}, a>0$, then the value of $\frac{\sqrt{x^{2}-1}}{x-\sqrt{x^{2}-1}}$ is:
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