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Question Number 148525 by nadovic last updated on 28/Jul/21

$$\mathrm{Express}\frac{\sqrt{15}+\sqrt{35}+\sqrt{21}+5}{\sqrt{3}+2\sqrt{5}+\sqrt{7}}\mathrm{in}\mathrm{the}\mathrm{form}$$$$a\sqrt{3}+b\sqrt{7}.$$

Answered by Ar Brandon last updated on 28/Jul/21

$$\frac{\sqrt{15}+\sqrt{35}+\sqrt{21}+5}{\sqrt{3}+2\sqrt{5}+\sqrt{7}}$$$$=\frac{\sqrt{3}\cdot \sqrt{5}+\sqrt{7}\cdot \sqrt{5}+\sqrt{3}\cdot \sqrt{7}+\sqrt{5}\stackrel{2}{}}{\sqrt{3}+2\sqrt{5}+\sqrt{7}}$$$$=\frac{\sqrt{5}\stackrel{2}{}+(\sqrt{3}+\sqrt{7})\sqrt{5}+\sqrt{7}\cdot \sqrt{5}}{\sqrt{3}+2\sqrt{5}+\sqrt{7}}$$$$=\frac{(\sqrt{5}+\sqrt{3})(\sqrt{5}+\sqrt{7})}{(\sqrt{5}+\sqrt{3})+(\sqrt{5}+\sqrt{7})}\times \frac{(\sqrt{5}+\sqrt{3})-(\sqrt{5}+\sqrt{7})}{(\sqrt{5}+\sqrt{3})-(\sqrt{5}+\sqrt{7})}$$

Answered by MJS_new last updated on 29/Jul/21

$$\frac{5+\sqrt{7}\sqrt{5}+\sqrt{7}\sqrt{3}+\sqrt{5}\sqrt{3}}{\sqrt{7}+2\sqrt{5}+\sqrt{3}}=$$$$=\frac{(5+\sqrt{7}\sqrt{5}+\sqrt{7}\sqrt{3}+\sqrt{5}\sqrt{3})(-\sqrt{7}+2\sqrt{5}+\sqrt{3})}{(\sqrt{7}+2\sqrt{5}+\sqrt{3})(-\sqrt{7}+2\sqrt{5}+\sqrt{3})}=$$$$=\frac{2\sqrt{7}\sqrt{5}\sqrt{3}+8\sqrt{7}+6\sqrt{5}+8\sqrt{3}}{16+4\sqrt{5}\sqrt{3}}=$$$$=\frac{\sqrt{7}\sqrt{5}\sqrt{3}+4\sqrt{7}+3\sqrt{5}+4\sqrt{3}}{2(4+\sqrt{5}\sqrt{3})}=$$$$=\frac{(\sqrt{7}\sqrt{5}\sqrt{3}+4\sqrt{7}+3\sqrt{5}+4\sqrt{3})(4-\sqrt{5}\sqrt{3})}{2(4+\sqrt{5}\sqrt{3})(4-\sqrt{5}\sqrt{3})}=$$$$=\frac{\sqrt{7}+\sqrt{3}}{2}$$$$\Rightarrow a=b=\frac{1}{2}$$

Commented by nadovic last updated on 30/Jul/21

$$ThankyouSir$$

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