[tex]\it a)\ \sqrt{12}-\sqrt{27}+\dfrac{1}{2}\sqrt{48}=\sqrt{4\cdot3}-\sqrt{9\cdot3}+\dfrac{1}{2}\cdot\sqrt{16\cdot3}=\\ \\ \\ =2\sqrt3-3\sqrt3+\dfrac{1}{2}\cdot4\sqrt3=2\sqrt3-3\sqrt3+2\sqrt3=4\sqrt3-3\sqrt3=\sqrt3\\ \\ \\ b)\ \dfrac{1}{3}\sqrt{18}+\dfrac{1}{2}\sqrt{8}-\dfrac{1}{5}\sqrt{50}=\dfrac{1}{3}\cdot3\sqrt{2}+\dfrac{1}{2}\cdot2\sqrt{2}-\dfrac{1}{5}\cdot5\sqrt{2}=\\ \\ \\ =\sqrt2+\sqrt2-\sqrt2=\sqrt2\\ \\ \\ c)\ (2\sqrt5+3\sqrt3)^2=(2\sqrt5)^2+2\cdot2\sqrt5\cdot3\sqrt2+(3\sqrt2)^2=\\ \\ \\ =20+12\sqrt{10}+18=38+12\sqrt{10}[/tex]
