Question : If $9\sqrt{x}=\sqrt{12}+\sqrt{147}$, then $x$ =?
Option 1: 5
Option 2: 3
Option 3: 2
Option 4: 4
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Correct Answer: 3
Solution : $9\sqrt{x} = \sqrt{12} + \sqrt{147}$ ⇒ $9\sqrt{x} = 2\sqrt{3} + 7\sqrt{3}$ ⇒ $9\sqrt{x} = 9\sqrt{3}$ $\therefore x = 3$ Hence, the correct answer is 3.
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Question : If $x=\frac{4\sqrt{15}}{\sqrt{5}+\sqrt{3}}$, the value of $\frac{x+\sqrt{20}}{x–\sqrt{20}}+\frac{x+\sqrt{12}}{x–\sqrt{12}}$ is:
Option 1: $1$
Option 2: $2$
Option 3: $\sqrt{3}$
Option 4: $\sqrt{5}$
Question : If $x^{4}+\frac{1}{x^{4}}=16$, then what is the value of $x^{2}+\frac{1}{x^{2}}$?
Option 1: $3 \sqrt{2}$
Option 2: $2 \sqrt{2}$
Option 3: $5 \sqrt{2 }$
Option 4: $4 \sqrt{2}$
Question : If $x=(7+3 \sqrt{5})$, then find the value of $x^2+\frac{1}{x^2}$.
Option 1: $ \frac{580+315 \sqrt{5}}{8}$
Option 2: $\frac{799+328 \sqrt{5}}{8}$
Option 3: $\frac{799+315 \sqrt{5}}{12}$
Option 4: $\frac{799+315 \sqrt{5}}{8}$
Question : If $x=(\sqrt{6}-1)^{\frac{1}{3}}$, then the value of $\left(x-\frac{1}{x}\right)^3+3\left(x-\frac{1}{x}\right)$ is:
Option 1: $\frac{2 \sqrt{6}-6}{5}$
Option 2: $\frac{4 \sqrt{6}-6}{5}$
Option 3: $\frac{4 \sqrt{6}-6}{3}$
Option 4: $\frac{4 \sqrt{3}-6}{5}$
Question : If $x=\frac{1}{x-5}(x>0)$, then the value of $x+\frac{1}{x}$ is:
Option 1: $\sqrt{41}$
Option 2: $\sqrt{29}$
Option 3: $\sqrt{23}$
Option 4: $\sqrt{43}$
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