Question

$\frac{\sqrt2}{\left(\sqrt6+2\right)}$ = কত?

Options :

• (A) $\sqrt 3 +\sqrt 2$
• (B) $2 -\sqrt 2$
• (C) $\sqrt 3 -\sqrt 2$
• (D) $\sqrt 3 +2$

Answer 1

$\frac{\sqrt2}{\left(\sqrt6+2\right)}$

= $\frac{\sqrt2\left(\sqrt6-2\right)}{\left(\sqrt6+2\right)\left(\sqrt6-2\right)}$ [ হর ও লব কে $\left(\sqrt6-2\right)$ দ্বারা গুণ করে ]

= $\frac{\sqrt2.\sqrt6-\sqrt2.2}{\left(\sqrt6\right)^2-\left(2\right)^2}$

= $\frac{\sqrt{2\times6}-2\sqrt2}{6-4}$

= $\frac{\sqrt{12}-2\sqrt2}2$

= $\frac{\sqrt{4\times3}-2\sqrt2}2$

= $\frac{\sqrt4.\sqrt3-2\sqrt2}2$

= $\frac{2\sqrt3-2\sqrt2}2$

= $\frac{2\left(\sqrt3-\sqrt2\right)}2$

= $\sqrt3-\sqrt2$