Verified by Ajay Kumar
Simplify: \(\frac{1}{\sqrt5-2}+\frac{1}{\sqrt5+2}\)
\(\,2\sqrt5\)
\(\sqrt6\)
\(\sqrt5\)
\(\,2\sqrt6\)
Answer: A
Taking the LCM, \(\frac{1}{\sqrt5-2}+\frac{1}{\sqrt5+2}=\frac{(\sqrt5+2)+(\sqrt5-2)}{(\sqrt5-2)(\sqrt5+2)}\). Using \((a-b)(a+b)=a^2-b^2\), this becomes \(\frac{2\sqrt5}{5-4}=2\sqrt5\). Trick: Use the difference of squares for conjugate denominators.
सरल करें: \(\frac{1}{\sqrt5-2}+\frac{1}{\sqrt5+2}\)
\(\,2\sqrt5\)
\(\sqrt6\)
\(\sqrt5\)
\(\,2\sqrt6\)
उत्तर: A
LCM लेने पर \(\frac{1}{\sqrt5-2}+\frac{1}{\sqrt5+2}=\frac{(\sqrt5+2)+(\sqrt5-2)}{(\sqrt5-2)(\sqrt5+2)}\) मिलता है। सूत्र \((a-b)(a+b)=a^2-b^2\) से यह \(\frac{2\sqrt5}{5-4}=2\sqrt5\) हो जाता है। ट्रिक: संयुग्मी हरों के लिए वर्गों के अंतर का सूत्र प्रयोग करें।
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