If $\frac{1}{\sqrt{7}-\sqrt{6}}$ is rationalised, the result is:

To clean up a fraction with a surd in the denominator, multiply numerator and denominator by something that makes the denominator rational.

**Single surd**: $\frac{1}{\sqrt{2}}$ → multiply by $\frac{\sqrt{2}}{\sqrt{2}}$ → $\frac{\sqrt{2}}{2}$.

**Sum/difference with surds**: $\frac{1}{3+\sqrt{2}}$ → multiply by the **conjugate** $(3-\sqrt{2}):\frac{1}{3+\sqrt{2}}\times \frac{3-\sqrt{2}}{3-\sqrt{2}}=\frac{3-\sqrt{2}}{(3{)}^2-(\sqrt{2}{)}^2}=\frac{3-\sqrt{2}}{9-2}=\frac{3-\sqrt{2}}{7}$.

The conjugate trick uses the difference-of-squares identity $(a+b)(a-b)=a^2-b^2$.