Question

Find the value of $\frac{1}{\sqrt{3}-\sqrt{2}-\sqrt{1}}$ if $\sqrt{2}=1.4$ and $\sqrt{6}=2.4$

Answer 1

$\frac{1}{\sqrt{3}-\sqrt{2}-\sqrt{1}}$

$=\frac{1}{\left(\sqrt{3}-\sqrt{2}\right)-1}\times \frac{\left(\sqrt{3}-\sqrt{2}\right)+1}{\left(\sqrt{3}-\sqrt{2}\right)+1}$

$=\frac{\sqrt{3}-\sqrt{2}+1}{{\left(\sqrt{3}-\sqrt{2}\right)}^2-1}$

$=\frac{\sqrt{3}-\sqrt{2}+1}{3+2-2\sqrt{6}-1}$

$=\frac{\sqrt{3}-\sqrt{2}+1}{4-2\sqrt{6}}$

$=\frac{\frac{\sqrt{6}}{\sqrt{2}}-\sqrt{2}+1}{4-2\sqrt{6}}$

$=\frac{12-9.8+7}{7\left(4-4.8\right)}$

$=\frac{19-9.8}{-5.6}$

$=-1.64$