Question

rationalize the denominator of 4/ 2 +√3 +√7

Answer 1

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Answer 2

Answer:

$\frac{4}{2+\sqrt{3}+\sqrt{7}}$= $\frac{(2\sqrt{3}+3-\sqrt{21})}{3}$

Explanation:

Given $\frac{4}{2+\sqrt{3}+\sqrt{7}}$

multiply numerator and denominator by (2+√3-√7), we get

=$\frac{4(2+\sqrt{3}-\sqrt7)}{[(2+\sqrt{3})+\sqrt{7}][(2+\sqrt{3})-\sqrt{3}]}$

= $\frac{4(2+\sqrt{3}-\sqrt{7})}{\left(2+\sqrt{3}\right)^{2}-\left(\sqrt{7}\right)^{2}}$

= $\frac{4(2+\sqrt{3}-\sqrt{7})}{2^{2}+2\times2\times\sqrt{3}+\left(\sqrt{3}\right)^{2}-7}$

=$\frac{4(2+\sqrt{3}-\sqrt{7})}{4+4\sqrt{3}+3-7}$

=$\frac{4(2+\sqrt{3}-\sqrt{7})}{4\sqrt{3}}$

=$\frac{(2+\sqrt{3}-\sqrt{7})}{\sqrt{3}}$

Rationalising the denominator, we get

= $\frac{(2+\sqrt{3}-\sqrt{7})\times\sqrt{3}}{\sqrt{3}\times\sqrt{3}}$

= $\frac{(2\sqrt{3}+3-\sqrt{21})}{3}$

Therefore,

$\frac{4}{2+\sqrt{3}+\sqrt{7}}$= $\frac{(2\sqrt{3}+3-\sqrt{21})}{3}$

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