যদি \(x=2, y=3\) এবং \(z=6\) হয়, তবে, \(\cfrac{3√x}{√y+√z}-\cfrac{4√y}{√z+√x}+\cfrac{√z}{√x+√y}\) -এর মান হিসাব করে লিখি ।
\(\cfrac{3√x}{√y+√z}-\cfrac{4√y}{√z+√x}+\cfrac{√z}{√x+√y}\)
\(=\cfrac{3\sqrt2}{\sqrt3+\sqrt6}-\cfrac{4\sqrt3}{\sqrt6+\sqrt2}+\cfrac{\sqrt6}{\sqrt2+\sqrt3}\)
\(=\cfrac{3\sqrt2}{\sqrt6+\sqrt3}-\cfrac{4\sqrt3}{\sqrt6+\sqrt2}+\cfrac{\sqrt6}{\sqrt3+\sqrt2}\)
\(=\cfrac{3\sqrt2(\sqrt6-\sqrt3)}{(\sqrt6+\sqrt3)(\sqrt6-\sqrt3)}-\cfrac{4\sqrt3(\sqrt6-\sqrt2)}{(\sqrt6+\sqrt2)(\sqrt6-\sqrt2)} \) \(+\cfrac{\sqrt6(\sqrt3-\sqrt2)}{(\sqrt3+\sqrt2)(\sqrt3-\sqrt2)}\)
\(=\cfrac{3\sqrt2(\sqrt6-\sqrt3)}{6-3}-\cfrac{4\sqrt3(\sqrt6-\sqrt2)}{6-2} \) \(+\cfrac{\sqrt6(\sqrt3-\sqrt2)}{3-2}\)
\(=\cfrac{\cancel3\sqrt2(\sqrt6-\sqrt3)}{\cancel3}-\cfrac{\cancel4\sqrt3(\sqrt6-\sqrt2)}{\cancel4}\)
\(+\cfrac{\sqrt6(\sqrt3-\sqrt2)}{1}\)
\(=\sqrt2(\sqrt6-\sqrt3)-\sqrt3(\sqrt6-\sqrt2)+\sqrt6(\sqrt3-\sqrt2)\)
\(=\cancel{2\sqrt3}-\cancel{\sqrt6}-\cancel{3\sqrt2}+\cancel{\sqrt6}+\cancel{3\sqrt2}-\cancel{2\sqrt3}\)
\(=0\)