Question

SHOW THAT 1 DIVIDED BY 1 + (ROOT)2

PLUS 1 DIVIDED BY (ROOT)2 +(ROOT)3 IS -1+(ROOT)3

Answer 1

The expression given to us is:

$$\begin{gathered} \frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{2}+\sqrt{3}} \\ \mathrm{To}\mathrm{show}:\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}+\sqrt{2}}=-1+\sqrt{3} \\ \mathrm{Consider}\mathrm{L}.\mathrm{H}.\mathrm{S}=\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}+\sqrt{2}} \\ \mathrm{Rationalising}\mathrm{we}\mathrm{get}, \\ \frac{1}{\sqrt{2}+1}\times \frac{\sqrt{2}-1}{\sqrt{2}-1}+\frac{1}{\sqrt{3}+\sqrt{2}}\times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}} \\ =\frac{\sqrt{2}-1}{{\left(\sqrt{2}\right)}^2-{\left(1\right)}^2}+\frac{\sqrt{3}-\sqrt{2}}{{\left(\sqrt{3}\right)}^2-{\left(\sqrt{2}\right)}^2} \\ =\frac{\sqrt{2}-1}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2} \\ =\frac{\sqrt{2}-1}{1}+\frac{\sqrt{3}-\sqrt{2}}{1} \\ =\frac{\sqrt{2}-1+\sqrt{3}-\sqrt{2}}{1} \\ =\sqrt{3}-1\mathrm{or}-1+\sqrt{3}=\mathrm{R}.\mathrm{H}.\mathrm{S}\left(\mathrm{answer}\right) \end{gathered}$$