Question : Let $a=\frac{1}{2-\sqrt{3}}+\frac{1}{3-\sqrt{8}}+\frac{1}{4-\sqrt{15}}$ then we have:
Option 1: $a<18 \text{ but } a\neq 9$
Option 2: $a>18$
Option 3: $a=18$
Option 4: $a=9$
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Correct Answer: $a<18 \text{ but } a\neq 9$
Solution :
$a=\frac{1}{2-\sqrt{3}}+\frac{1}{3-\sqrt{8}}+\frac{1}{4-\sqrt{15}}$
$=\frac{1}{2-\sqrt{3}}\times\frac{2+\sqrt{3}}{2+\sqrt{3}}+\frac{1}{3-\sqrt{8}}\times\frac{3+\sqrt{8}}{3+\sqrt{8}}+\frac{1}{4-\sqrt{15}}\times\frac{4+\sqrt{15}}{4+\sqrt{15}}$
$=\frac{2+\sqrt{3}}{4-3}+\frac{3+\sqrt{8}}{9-8}+\frac{4+\sqrt{15}}{16-15}$
$=2+\sqrt{3}+3+\sqrt{8}+4+\sqrt{15}$
$=9+\sqrt{3}+\sqrt{8}+\sqrt{15}$
$<9+2+3+4$
$<18$
Hence, the correct answer is $a<18 \text{ but } a\neq 9$.
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