Question

Show that: 1/3-root 8 - 1/ root 8-root 7 + 1/root 7-root 6 - 1/ root 6-root 5 + 1/ root 5-2 = 5

Answer 1

$$\begin{gathered} \frac{1}{3-\sqrt{8}}-\frac{1}{\sqrt{8}-\sqrt{7}}+\frac{1}{\sqrt{7}-\sqrt{6}}-\frac{1}{\sqrt{6}-\sqrt{5}}+\frac{1}{\sqrt{5}-2} \\ =\frac{1}{3-\sqrt{8}}\times \frac{3+\sqrt{8}}{3+\sqrt{8}}-\frac{1}{\sqrt{8}-\sqrt{7}}\times \frac{\sqrt{8}+\sqrt{7}}{\sqrt{8}+\sqrt{7}}+\frac{1}{\sqrt{7}-\sqrt{6}}\times \frac{\sqrt{7}+\sqrt{6}}{\sqrt{7}+\sqrt{6}}-\frac{1}{\sqrt{6}-\sqrt{5}}\times \frac{\sqrt{6}+\sqrt{5}}{\sqrt{6}+\sqrt{5}}+\frac{1}{\sqrt{5}-2}\times \frac{\sqrt{5}+2}{\sqrt{5}+2} \\ =\frac{3+\sqrt{8}}{{\left(3\right)}^2-{\left(\sqrt{8}\right)}^2}-\frac{\left(\sqrt{8}+\sqrt{7}\right)}{{\left(\sqrt{8}\right)}^2-{\left(\sqrt{7}\right)}^2}+\frac{\sqrt{7}+\sqrt{6}}{{\left(\sqrt{7}\right)}^2-{\left(\sqrt{6}\right)}^2}-\frac{\left(\sqrt{6}+\sqrt{5}\right)}{{\left(\sqrt{6}\right)}^2-{\left(\sqrt{5}\right)}^2}+\frac{\sqrt{5}+2}{{\left(\sqrt{5}\right)}^2-{\left(2\right)}^2} \\ =\frac{3+\sqrt{8}}{9-8}-\frac{\left(\sqrt{8}+\sqrt{7}\right)}{8-7}+\frac{\sqrt{7}+\sqrt{6}}{7-6}-\frac{\left(\sqrt{6}+\sqrt{5}\right)}{6-5}+\frac{\sqrt{5}+2}{5-4} \\ =\frac{3+\sqrt{8}}{1}-\frac{\left(\sqrt{8}+\sqrt{7}\right)}{1}+\frac{\sqrt{7}+\sqrt{6}}{1}-\frac{\left(\sqrt{6}+\sqrt{5}\right)}{1}+\frac{\sqrt{5}+2}{1} \\ =3+\sqrt{8}-\left(\sqrt{8}+\sqrt{7}\right)+\sqrt{7}+\sqrt{6}-\left(\sqrt{6}+\sqrt{5}\right)+\sqrt{5}+2 \\ =3+\sqrt{8}-\sqrt{8}-\sqrt{7}+\sqrt{7}+\sqrt{6}-\sqrt{6}-\sqrt{5}+\sqrt{5}+2 \\ =3+2 \\ =5 \end{gathered}$$