Question

Rationalise the denominator 2/ root 5+root 3+1/root 3 +root2 -3/root 5+root 2

Answer 1

hope this helps you....

Answer 2

Answer:

$\frac{2}{\sqrt{5}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{2}}-\frac{3}{\sqrt{5}+\sqrt{2}}\\=0$

Step-by-step explanation:

$i) \frac{2}{\sqrt{5}+\sqrt{3}}\\=\frac{2(\sqrt{5}-\sqrt{3})}{(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})}\\=\frac{2(\sqrt{5}-\sqrt{3})}{(\sqrt{5})^{2}-(\sqrt{3})^{2}}\\=</p><p>\frac{2(\sqrt{5}-\sqrt{3})}{(5-3)}\\=\frac{2(\sqrt{5}-\sqrt{3})}{2}\\=\sqrt{5}-\sqrt{3}---(1)$

$ii) \frac{1}{\sqrt{3}+\sqrt{2}}\\=\frac{(\sqrt{3}-\sqrt{2})}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}\\=\frac{(\sqrt{3}-\sqrt{2})}{(\sqrt{3})^{2}-(\sqrt{2})^{2}}\\=</p><p>\frac{(\sqrt{3}-\sqrt{2})}{(3-2)}\\=\frac{(\sqrt{3}-\sqrt{2})}{1}\\=\sqrt{3}-\sqrt{2}---(2)$

$iii) \frac{3}{\sqrt{5}+\sqrt{2}}\\=\frac{3(\sqrt{5}-\sqrt{2})}{(\sqrt{5}+\sqrt{2})(\sqrt{5}-\sqrt{2})}\\=\frac{3(\sqrt{5}-\sqrt{2})}{(\sqrt{5})^{2}-(\sqrt{2})^{2}}\\=</p><p>\frac{3(\sqrt{5}-\sqrt{2})}{(5-2)}\\=\frac{3(\sqrt{5}-\sqrt{2})}{3}\\=\sqrt{5}-\sqrt{2}---(3)$

$Now,\\\frac{2}{\sqrt{5}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{2}}-\frac{3}{\sqrt{5}+\sqrt{2}}\\=\sqrt{5}-\sqrt{3}+\sqrt{3}-\sqrt{2}-(\sqrt{5}-\sqrt{2})\\ [from \\(1),(2)\:and\:(3)]$

$=\sqrt{5}-\sqrt{3}+\sqrt{3}-\sqrt{2}-\sqrt{5}+\sqrt{2}\\=0$

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