If $\sqrt{3}=1.732$ and $\sqrt{2}=1.414$, then find the value of $\frac{1}{\sqrt{3}-\sqrt{2}}$.

Given: $\sqrt{3}=1.732, \sqrt{2}=1.414$

Suppose $t=\frac{1}{\sqrt{3}-\sqrt{2}}$

Let's rationalize by multiplying both numerator and denominator by $\sqrt{3}+\sqrt{2}$

$t=\frac{1}{\sqrt{3}-\sqrt{2}}\times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}$

$t=\frac{\sqrt{3}+\sqrt{2}}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}$

$t=\frac{\sqrt{3}+\sqrt{2}}{(\sqrt{3}{)}^2-(\sqrt{2}{)}^2}$ [Since $(a+b)(a-b)=a^2-b^2$]

$t=\frac{\sqrt{3}+\sqrt{2}}{3-2}$

$t=\frac{\sqrt{3}+\sqrt{2}}{1}$

$t=\sqrt{3}+\sqrt{2}$ -----(1)

Now put the given values in equation (1)

$t=(1.732)+(1.414)$

$t=3.146$
Here, we get  $\frac{1}{\sqrt{3}-\sqrt{2}}=3.146$.