Question : If $x+ \sqrt{5} = 5+\sqrt{y}$ are positive integers, then the value of $\frac{\sqrt{x}+y}{x+ \sqrt{y}}$ is:
Option 1: 1
Option 2: 2
Option 3: $\sqrt{5}$
Option 4: 5
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Correct Answer: 1
Solution : Given: $x+ \sqrt{5} = 5+\sqrt{y}$ By comparing rational and irrational numbers, we get ⇒ $x = 5$ and $\sqrt{y} = \sqrt{5}$ ⇒ $y = 5$ By putting the values of $x$ and $y$ in the given expression, $ \frac{\sqrt{x}+y}{x+ \sqrt{y}}$ $=\frac{\sqrt{5}+5}{5+ \sqrt{5}}$ $= 1$ Hence, the correct answer is 1.
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Question : If $x=\frac{\sqrt{5}+1}{\sqrt{5}-1}$ and $y=\frac{\sqrt{5}-1}{\sqrt{5}+1}$, then the value of $\frac{x^{2}+xy+y^{2}}{x^{2}-xy+y^{2}}$ is:
Question : If $x=\frac{\sqrt{5}-\sqrt{4}}{\sqrt{5}+\sqrt{4}}$ and $y=\frac{\sqrt{5}+\sqrt{4}}{\sqrt{5}-\sqrt{4}}$ then the value of $\frac{x^2-x y+y^2}{x^2+x y+y^2}=$?
Question : If $\frac{\sqrt{5+x}+\sqrt{5-x}}{\sqrt{5+x}-\sqrt{5-x}}=3$, what is the value of $x$?
Question : If $\sin17°=\frac{x}{y}$, then the value of $(\sec17°–\sin73°)$ is:
Question : If $x^4+y^4+x^2 y^2=17 \frac{1}{16}$ and $x^2-x y+y^2=5 \frac{1}{4}$, then one of the values of $(x-y)$ is:
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