যদি \(x = \cfrac{4\sqrt{15}}{\sqrt5+\sqrt3}\) হয়, তবে \(\cfrac{x+\sqrt{20}}{x-\sqrt{20}}+\cfrac{x+\sqrt{12}}{x-\sqrt{12}}\) এর মান নির্ণয় করো । Madhyamik 2025
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\(x = \cfrac{4\sqrt{15}}{\sqrt5+\sqrt3}\)
\(=\cfrac{4\sqrt{15}(\sqrt5-\sqrt3)}{(\sqrt5+\sqrt3)(\sqrt5-\sqrt3)}\)
\(=\cfrac{4\sqrt{15}(\sqrt5-\sqrt3)}{(\sqrt5)^2-(\sqrt3)^2}\)
\(=\cfrac{4\sqrt{15}(\sqrt5-\sqrt3)}{5-3}\)
\(=\cfrac{\cancel42\sqrt{15}(\sqrt5-\sqrt3)}{\cancel2}\)
\(=2\sqrt{3\times 5\times 5}-2\sqrt{3\times 5\times 3}\)
\(=10\sqrt3-6\sqrt5\)

\(\therefore \cfrac{x+\sqrt{20}}{x-\sqrt{20}}=\cfrac{10\sqrt3-6\sqrt5+\sqrt{20}}{10\sqrt3-6\sqrt5-\sqrt{20}}\)
\(=\cfrac{10\sqrt3-6\sqrt5+2\sqrt{5}}{10\sqrt3-6\sqrt5-2\sqrt{5}}\)
\(=\cfrac{10\sqrt3-4\sqrt5}{10\sqrt3-8\sqrt5}\)
\(=\cfrac{5\sqrt3-2\sqrt5}{5\sqrt3-4\sqrt5}\)
\(=\cfrac{(5\sqrt3-2\sqrt5)(5\sqrt3+4\sqrt5)}{(5\sqrt3-4\sqrt5)(5\sqrt3+4\sqrt5)}\)
\(=\cfrac{75+20\sqrt{15}-10\sqrt{15}-40}{75-80}\)
\(=\cfrac{35+10\sqrt{15}}{-5}\)
\(=-7-2\sqrt{15}\)

এবং \(\cfrac{x+\sqrt{12}}{x-\sqrt{12}}=\cfrac{10\sqrt3-6\sqrt5+\sqrt{12}}{10\sqrt3-6\sqrt5-\sqrt{12}}\)
\(=\cfrac{10\sqrt3-6\sqrt5+2\sqrt{3}}{10\sqrt3-6\sqrt5-2\sqrt{3}}\)
\(=\cfrac{12\sqrt3-6\sqrt5}{8\sqrt3-6\sqrt5}\)
\(=\cfrac{6\sqrt3-3\sqrt5}{4\sqrt3-3\sqrt5}\)
\(=\cfrac{(6\sqrt3-3\sqrt5)(4\sqrt3+3\sqrt5)}{(4\sqrt3-3\sqrt5)(4\sqrt3+4\sqrt3)}\)
\(=\cfrac{72+18\sqrt{15}-12\sqrt{15}-45}{48-45}\)
\(=\cfrac{27+6\sqrt{15}}{3}\)
\(=9+2\sqrt{15}\)

এখন \(\cfrac{x+\sqrt{20}}{x-\sqrt{20}}+\cfrac{x+\sqrt{12}}{x-\sqrt{12}}\)
\(=-7-2\sqrt{15}+9+2\sqrt{15}\)
\(=2\) [Answer]

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