x=17−16,y=15−14⇒

Given x=17−16,y=15−14x=17−16=(17−16)×(17+16)(17+16)=17−1617+16=117+16y=15−14=(15−14)×(15+14)(15+14)=115+14since 15+14 117+16⇒ y>x⇒x<y

x=17−16,y=15−14⇒

Given x=17−16,y=15−14x=17−16=(17−16)×(17+16)(17+16)=17−1617+16=117+16y=15−14=(15−14)×(15+14)(15+14)=115+14since 15+14 117+16⇒ y>x⇒x<y

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$x = \sqrt{17} - \sqrt{16}, y = \sqrt{15} - \sqrt{14} \Rightarrow$

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x > y

x < y

x = y

xy = 1

answer is C.

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Detailed Solution

Given $x = \sqrt{17} - \sqrt{16}, y = \sqrt{15} - \sqrt{14}$
$x = \sqrt{17} - \sqrt{16} = (\sqrt{17} - \sqrt{16}) \times \frac{(\sqrt{17} + \sqrt{16})}{(\sqrt{17} + \sqrt{16})} = \frac{17 - 16}{\sqrt{17} + \sqrt{16}} = \frac{1}{\sqrt{17} + \sqrt{16}}$

$y = \sqrt{15} - \sqrt{14} = (\sqrt{15} - \sqrt{14}) \times \frac{(\sqrt{15} + \sqrt{14})}{(\sqrt{15} + \sqrt{14})} = \frac{1}{\sqrt{15} + \sqrt{14}}$
since $\sqrt{15} + \sqrt{14} < \sqrt{17} + \sqrt{16}$
$\begin{array}{l} \Rightarrow \frac{1}{\sqrt{15} + \sqrt{14}} > \frac{1}{\sqrt{17} + \sqrt{16}} \\ \Rightarrow y > x \\ \Rightarrow x < y \end{array}$