Question

simplify 3 root 2 upon root 6 minus under root 3 + 2 root 2 upon root 6 + root 2 minus 4 root 3 upon under root 6 minus under root 2​

Answer 1

$\underline{\textsf{Given:}}$

$\dfrac{3\sqrt{2}}{\sqrt{6}-\sqrt{3}}+\dfrac{2\sqrt{3}}{\sqrt{6}+\sqrt{2}}-\dfrac{4\sqrt{3}}{\sqrt{6}-\sqrt{2}}$

$\underline{\textsf{To find:}}$

$\textsf{Simplified form of the given expression}$

$\underline{\textsf{Solution:}}$

$\textsf{Consider,}$

$\dfrac{3\sqrt{2}}{\sqrt{6}-\sqrt{3}}+\dfrac{2\sqrt{3}}{\sqrt{6}+\sqrt{2}}-\dfrac{4\sqrt{3}}{\sqrt{6}-\sqrt{2}}$

$\textsf{First we rationalize denominator of each fraction}$

$\dfrac{3\sqrt{2}}{\sqrt{6}-\sqrt{3}}$

$=\dfrac{3\sqrt{2}}{\sqrt{6}-\sqrt{3}}{\times}\dfrac{\sqrt{6}+\sqrt{3}}{\sqrt{6}+\sqrt{3}}$

$=\dfrac{3\sqrt{2}(\sqrt{6}+\sqrt{3})}{6-3}$

$=\dfrac{3\sqrt{2}(\sqrt{6}+\sqrt{3})}{3}$

$=\sqrt{2}(\sqrt{6}+\sqrt{3})$

$=\sqrt{12}+\sqrt{6}$......(1)

$\dfrac{2\sqrt{3}}{\sqrt{6}+\sqrt{2}}$

$=\dfrac{2\sqrt{3}}{\sqrt{6}+\sqrt{2}}{\times}\dfrac{\sqrt{6}-\sqrt{2}}{\sqrt{6}-\sqrt{2}}$

$=\dfrac{2\sqrt{3}(\sqrt{6}-\sqrt{2})}{6-2}$

$=\dfrac{2\sqrt{3}(\sqrt{6}-\sqrt{2})}{4}$

$=\dfrac{\sqrt{3}(\sqrt{6}-\sqrt{2})}{2}$

$=\dfrac{\sqrt{18}-\sqrt{6}}{2}$.....(2)

$\dfrac{4\sqrt{3}}{\sqrt{6}-\sqrt{2}}$

$=\dfrac{4\sqrt{3}}{\sqrt{6}-\sqrt{2}}{\times}\dfrac{\sqrt{6}+\sqrt{2}}{\sqrt{6}+\sqrt{2}}$

$=\dfrac{4\sqrt{3}(\sqrt{6}+\sqrt{2})}{6-2}$

$=\dfrac{4\sqrt{3}(\sqrt{6}+\sqrt{2})}{4}$

$=\sqrt{3}(\sqrt{6}+\sqrt{2})$

$=\sqrt{18}+\sqrt{6}$....(3)

$\text{Now,}$

$\dfrac{3\sqrt{2}}{\sqrt{6}-\sqrt{3}}+\dfrac{2\sqrt{3}}{\sqrt{6}+\sqrt{2}}-\dfrac{4\sqrt{3}}{\sqrt{6}-\sqrt{2}}$

$=\sqrt{12}+\sqrt{6}+\dfrac{\sqrt{18}-\sqrt{6}}{2}-(\sqrt{18}+\sqrt{6})$

$=\sqrt{12}+\sqrt{6}+\dfrac{\sqrt{18}-\sqrt{6}}{2}-\sqrt{18}-\sqrt{6}$

$=\sqrt{12}+\dfrac{\sqrt{18}-\sqrt{6}}{2}-\sqrt{18}$

$=\dfrac{2\sqrt{12}+\sqrt{18}-\sqrt{6}-2\sqrt{18}}{2}$

$=\dfrac{2\sqrt{12}-\sqrt{6}-\sqrt{18}}{2}$

$=\dfrac{4\sqrt{3}-\sqrt{6}-3\sqrt{2}}{2}$

$\underline{\textsf{Answer:}}$

$\dfrac{3\sqrt{2}}{\sqrt{6}-\sqrt{3}}+\dfrac{2\sqrt{3}}{\sqrt{6}+\sqrt{2}}-\dfrac{4\sqrt{3}}{\sqrt{6}-\sqrt{2}}=\dfrac{4\sqrt{3}-\sqrt{6}-3\sqrt{2}}{2}$

Answer 2

Answer:

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