If $x=2+\sqrt{3},$ then find the value of $x^{2}+\frac{1}{x^{2}}$ and $x^{3}+\frac{1}{x^{3}}$

Rationalise the denominator in each of the following and hence evaluate by taking $\sqrt{2}=1.414, \sqrt{3}=1.732$ and $\sqrt{5}=2.236,$ upto three places of decimal.

$\frac{\sqrt{2}}{2+\sqrt{2}}$

Which of the following is equal to $x$?

A rational number between $\sqrt{2}$ and $\sqrt{3}$ is

Rationalise the denominator of the following:

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

Prove that

$\left(\frac{x^{a}}{x^{b}}\right)^{a+b} \times\left(\frac{x^{b}}{x^{c}}\right)^{b+c} \times\left(\frac{x^{c}}{x^{a}}\right)^{c+a}=1$