Question

rationalise the denominator 14/5√3 - √5

Answer 1

Your answer is attached.

Answer 2

Answer:

$\frac{14}{5\sqrt{3}-\sqrt{5}}$

=$\frac{5\sqrt{3}+\sqrt{5}}{5}$

Step-by-step explanation:

Given $\frac{14}{5\sqrt{3}-\sqrt{5}}$

Multiply numerator and denominator by $5\sqrt{3}+\sqrt{5}$ ,we get

=$\frac{14(5\sqrt{3}+\sqrt{5})}{(5\sqrt{3}-\sqrt{5})(5\sqrt{3}+\sqrt{5})}$

=$\frac{14(5\sqrt{3}+\sqrt{5})}{\left(5\sqrt{3}\right)^{2}-\left(\sqrt{5}\right)^{2}}$

/* By algebraic identity:

(a+b)(a-b)=a²-b² */

=$\frac{14(5\sqrt{3}+\sqrt{5})}{75-5}$

= $\frac{14(5\sqrt{3}+\sqrt{5})}{70}$

After cancellation, we get

= $\frac{5\sqrt{3}+\sqrt{5}}{5}$

Therefore,

$\frac{14}{5\sqrt{3}+\sqrt{5}}$

= $\frac{5\sqrt{3}+\sqrt{5}}{5}$

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