A) \[\frac{63}{61}\]
B) \[\frac{67}{65}\]
C) \[\frac{65}{63}\]
D) \[\frac{69}{67}\]
Correct Answer: A
Solution :
(a): \[a=\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}},b=\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}\,\,\,\therefore a+b=\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}\] \[=\frac{{{\left( \sqrt{5}-\sqrt{3} \right)}^{2}}{{\left( \sqrt{5}+\sqrt{3} \right)}^{2}}}{\left( \sqrt{5}+\sqrt{3} \right)\times \left( \sqrt{5}-\sqrt{3} \right)}=\frac{2\left( {{\left( \sqrt{5} \right)}^{2}}+{{\left( \sqrt{3} \right)}^{2}} \right)}{5-3}=5+3=8\] \[ab=\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\times \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}=1\] \[\therefore \frac{{{a}^{2}}-ab+{{b}^{2}}}{{{a}^{2}}+ab+{{b}^{2}}}=\frac{{{\left( a+b \right)}^{2}}-3ab}{{{\left( a+b \right)}^{2}}-3ab}\] \[=\frac{{{8}^{2}}-1}{{{8}^{2}}-3}=\frac{63}{61}\]
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