Question

rationalise the denominator 3√2/√6-√3 -4√3/√6-√2+2√3/2+√6​

Answer 1

Answer:

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

Step-by-step explanation:

$3\cdot \frac{\sqrt{2}}{\sqrt{6}}-\sqrt{3}-4\cdot \frac{\sqrt{3}}{\sqrt{6}}-\sqrt{2}+2\cdot \frac{\sqrt{3}}{2}+\sqrt{6}$

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

$\mathrm{Add\:similar\:elements:}\:-2\sqrt{2}-\sqrt{2}=-3\sqrt{2}$

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

$\mathrm{Add\:similar\:elements:}\:\sqrt{3}-\sqrt{3}+\sqrt{3}=\sqrt{3}$

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

Answer 2

above answer is correct

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