find the number obtained on rationalising the denominator of 2/root11 - 3

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Solution

Dear student
$\frac{2}{\sqrt{11} - 3} = \frac{2}{\sqrt{11} - 3} \times \frac{\sqrt{11} + 3}{\sqrt{11} + 3} = \frac{2 (\sqrt{11} + 3)}{11 - 3} = \frac{2 (\sqrt{11} + 3)}{8}$
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Q. The number obtained on rationalising the denominator of

\frac{1}{\sqrt{9} - \sqrt{8}}

is ?

Q. Question 12
The number obtained on rationalising the denominator of

\frac{1}{\sqrt{7} - 2}

is

A)

\frac{\sqrt{7} + 2}{3}

\frac{\sqrt{7} - 2}{3}

\frac{\sqrt{7} + 2}{5}

\frac{\sqrt{7} + 2}{45}

Question 12
The number obtained on rationalising the denominator of $\frac{1}{\sqrt{7} - 2}$ is

A) $\frac{\sqrt{7} + 2}{3}$
B) $\frac{\sqrt{7} - 2}{3}$
C) $\frac{\sqrt{7} + 2}{5}$
D) $\frac{\sqrt{7} + 2}{45}$

Simplify the following by rationalising the denominator $\frac{6 - 4 \sqrt{3}}{6 + 4 \sqrt{3}} .$

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find the number obtained on rationalising the denominator of 2/root11 - 3

Dear student 211-3=211-3×11+311+3=211+311-3=211+38 Regards

Dear student
$\frac{2}{\sqrt{11} - 3} = \frac{2}{\sqrt{11} - 3} \times \frac{\sqrt{11} + 3}{\sqrt{11} + 3} = \frac{2 (\sqrt{11} + 3)}{11 - 3} = \frac{2 (\sqrt{11} + 3)}{8}$
Regards