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 $x=\frac{4\sqrt{ab}}{\sqrt a+ \sqrt b}$, then what is the value of $\frac{x+2\sqrt{a}}{x-2\sqrt a}+\frac{x+2\sqrt{b}}{x-2\sqrt b}$(when $a\neq b$)?
Option 1: 0
Option 2: 2
Option 3: 4
Option 4: $\frac{(\sqrt a+\sqrt b)}{(\sqrt a - \sqrt b)}$
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:
Option 1: $a+1$
Option 2: $\frac{1}{2}(a+1)$
Option 3: $\frac{1}{2}(a–1)$
Option 4: $a–1$
Question : If $c+ \frac{1}{c} =\sqrt{3}$, then the value of $c^{3}+ \frac{1}{c^{3}}$ is equal to:
Option 2: $\sqrt[3]{3}$
Option 3: $\frac{1}{\sqrt{3}}$
Option 4: $\sqrt[6]{3}$
Question : The value of $\frac{1}{4-\sqrt{15}}-\frac{1}{\sqrt{15}-\sqrt{14}}+\frac{1}{\sqrt{14}-\sqrt{13}}-\frac{1}{\sqrt{13}-\sqrt{12}}+\frac{1}{\sqrt{12}-\sqrt{11}}-\frac{1}{\sqrt{11}-\sqrt{10}}+\frac{1}{\sqrt{10}-3}-\frac{1}{3-\sqrt{8}}$ is:
Option 1: $2-2 \sqrt{2}$
Option 2: $4+2 \sqrt{2}$
Option 3: $4-2 \sqrt{2}$
Option 4: $2+2 \sqrt{2}$
Question : If $a=\frac{1}{a - 5}(a>0)$, then the value of $a+\frac{1}{a}$ is:
Option 1: $\sqrt{29}$
Option 2: $–\sqrt{27}$
Option 3: $-\sqrt{29}$
Option 4: $\sqrt{27}$
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