Aus Online Mathematik Brückenkurs 1

The root sign in the denominator lies in a quadratic term and we therefore expand first the quadratic \displaystyle \left( \sqrt{3}-2 \right)^{2} using the square rule

\displaystyle \begin{align} & \left( \sqrt{3}-2 \right)^{2}=\left( \sqrt{3} \right)^{2}-2\centerdot \sqrt{3}\centerdot 2+2^{2} \\ & =3-4\sqrt{3}+4=7-4\sqrt{3} \\ \end{align}

Thus,

\displaystyle \begin{align} & \left( \sqrt{3}-2 \right)^{2}=\left( \sqrt{3} \right)^{2}-2\centerdot \sqrt{3}\centerdot 2+2^{2} \\ & =3-4\sqrt{3}+4=7-4\sqrt{3} \\ & \frac{1}{\left( \sqrt{3}-2 \right)^{2}-2}=\frac{1}{7-4\sqrt{3}-2}=\frac{1}{5-4\sqrt{3}} \\ \end{align}

and in the expression we can get rid of the root sign from the denominator by multiplying the top and bottom of the equation by the conjugate \displaystyle 5+4\sqrt{3},

\displaystyle \begin{align} & \frac{1}{5-4\sqrt{3}}=\frac{1}{5-4\sqrt{3}}\centerdot \frac{5+4\sqrt{3}}{5+4\sqrt{3}}=\frac{5+4\sqrt{3}}{5^{2}-\left( 4\sqrt{3} \right)^{2}} \\ & =\frac{5+4\sqrt{3}}{5^{2}-4^{2}\left( \sqrt{3} \right)^{2}}=\frac{5+4\sqrt{3}}{25-16\centerdot 3} \\ & =\frac{5+4\sqrt{3}}{-23}=-\frac{5+4\sqrt{3}}{23} \\ \end{align}