Rationalise the denominator in each of the following:
$\frac{3+2 \sqrt{2}}{3-2 \sqrt{2}}$

$(17+12 \sqrt{2})$ To rationalise the denominator,multiply the numerator and the denominator by the conjugate of the denominator,which is $(3+2 \sqrt{2})$.
$\frac{3+2 \sqrt{2}}{3-2 \sqrt{2}} \times \frac{3+2 \sqrt{2}}{3+2 \sqrt{2}}$
$= \frac{(3+2 \sqrt{2})^2}{(3)^2 - (2 \sqrt{2})^2}$
$= \frac{3^2 + (2 \sqrt{2})^2 + 2(3)(2 \sqrt{2})}{9 - (4 \times 2)}$
$= \frac{9 + 8 + 12 \sqrt{2}}{9 - 8}$
$= \frac{17 + 12 \sqrt{2}}{1}$
$= 17 + 12 \sqrt{2}$