$\frac{\sqrt{\mathrm{75}}}{\sqrt{\mathrm{18}}-\sqrt{\mathrm{12}}}$

Express the inside as a product of a perfect square x a number

Perfect squares are: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, etc.

Their square roots is a number which when multiplied by itself gives the perfect square.

75 = 25 x 3

$\sqrt{\mathrm{75}}$ = $\sqrt{\mathrm{25\; x\; 3}}$ = $\sqrt{\mathrm{25}}$ x $\sqrt{3}$ = 5$\sqrt{3}$

18 = 9 x 2

$\sqrt{\mathrm{18}}$ = $\sqrt{\mathrm{9\; x\; 2}}$ = $\sqrt{9}$ x $\sqrt{2}$ = 3$\sqrt{2}$

12 = 4 x 3

$\sqrt{\mathrm{12}}$ = $\sqrt{\mathrm{4\; x\; 3}}$ = $\sqrt{4}$ x $\sqrt{3}$ = 2$\sqrt{3}$

Substitute the above into the original expression:

$\frac{5\sqrt{3}}{3\sqrt{2}-2\sqrt{3}}$

Rationalise the denominator by multiplying the numerator and denominator by the conjugate of the denominator.

The conjugate is simply changing the sign between the two terms.

Conjugate of (3$\sqrt{2}$ - 2$\sqrt{3}$) is (3$\sqrt{2}$ + 2$\sqrt{3}$)

$\frac{5\sqrt{3}}{3\sqrt{2}-2\sqrt{3}}$ = $\frac{5\sqrt{3}}{3\sqrt{2}-2\sqrt{3}}$ x $\frac{3\sqrt{2}+2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}$

Multiplying the denominator by it's conjugate produces a difference of two squares:

(3$\sqrt{2}$ - 2$\sqrt{3}$)(3$\sqrt{2}$ + 2$\sqrt{3}$) = (3$\sqrt{2}$)² - (2$\sqrt{3}$)² = 18 - 12 = 6

Notes:

1. The square affects each term inside the parentheses.
2. (3$\sqrt{2}$)² is 3 squared and $\sqrt{2}$ squared.
3. 3 squared is 9 and $\sqrt{2}$ squared is 2.
4. 9 x 2 = 18.
5. 2$\sqrt{3}$ squared is 2 squared and $\sqrt{3}$ squared.
6. 2 squared is 4 and $\sqrt{3}$ squared is 3.
7. 4 x 3 = 12.
8. When you square a square root, you get rid of the square root sign.

Multiplying the numerator:

5$\sqrt{3}$(3$\sqrt{2}$ + 2$\sqrt{3}$) = 5$\sqrt{3}$ x 3$\sqrt{2}$ + 5$\sqrt{3}$ x 2$\sqrt{3}$

5$\sqrt{3}$(3$\sqrt{2}$ + 2$\sqrt{3}$) = 5 x 3 x $\sqrt{\mathrm{3\; x\; 2}}$ + 5 x 2 x $\sqrt{\mathrm{3\; x\; 3}}$

5$\sqrt{3}$(3$\sqrt{2}$ + 2$\sqrt{3}$) = 15$\sqrt{6}$ + 10$\sqrt{9}$

But $\sqrt{9}$ = 3, so:

5$\sqrt{3}$(3$\sqrt{2}$ + 2$\sqrt{3}$) = 15$\sqrt{6}$ + 3 x 10 = 15$\sqrt{6}$ + 30

Putting it all together:

$\frac{15\sqrt{6}+30}{6}$ = $\frac{15\sqrt{6}+30}{6}$ = $\frac{15\sqrt{6}}{6}$ + $\frac{30}{6}$ = $\frac{5}{2}\sqrt{6}$ + 5

$\frac{\sqrt{\mathrm{75}}}{\sqrt{\mathrm{18}}-\sqrt{\mathrm{12}}}$ = 5 + $\frac{5}{2}\sqrt{6}$