Question

rationalise the denominator of the following. 1/root7-root6

Answer 1

See this attachment!!

Answer 2

Answer:

$\sqrt{7}+\sqrt{6}$

Explanation:

We have ,

$\frac{1}{\sqrt{7}-\sqrt{6}}$

Rationalising the denominator, we get

=$\frac{(\sqrt{7}+\sqrt{6})}{(\sqrt{7}-\sqrt{6})(\sqrt{7}+\sqrt{6}}$

= $\frac{(\sqrt{7}+\sqrt{6})}{(\sqrt{7})^{2}-(\sqrt{6})^{2}}$

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We know the algebraic identity:

a²-b² = (a+b)(a-b)

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= $\frac{(\sqrt{7}+\sqrt{6})}{(7-6)}$

= $\frac{(\sqrt{7}+\sqrt{6})}{1}$

= $\sqrt{7}+\sqrt{6}$

Therefore,

$\frac{1}{(\sqrt{7}-\sqrt{6})}= \sqrt{7}+\sqrt{6}$

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