Question

rationalise the denominator 6 upon root 12 minus root 3

Answer 1

here is your ans.
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Answer 2

The required expression is $\frac{6}{\sqrt{12}-\sqrt{3}}=\frac{13\sqrt{3}}{9}$.

Step-by-step explanation:

Given : Expression $\frac{6}{\sqrt{12}-\sqrt{3}}$

To find : Rationalize the denominator ?

Solution :

Expression $\frac{6}{\sqrt{12}-\sqrt{3}}$

Rationalize the denominator,

$=\frac{6}{\sqrt{12}-\sqrt{3}}\times \frac{\sqrt{12}+\sqrt{3}}{\sqrt{12}+\sqrt{3}}$

$=\frac{6(\sqrt{12}+\sqrt{3})}{(\sqrt{12})^2-(\sqrt{3})^2}$

$=\frac{12\sqrt{3}+\sqrt{3}}{12-3}$

$=\frac{13\sqrt{3}}{9}$

Therefore, the required expression is $\frac{6}{\sqrt{12}-\sqrt{3}}=\frac{13\sqrt{3}}{9}$.

#Learn more

Rationalise the denominator 6/9root3

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