Odpowiedź :
Obliczenia od a) do c)
[tex]\sqrt2\cdot\sqrt6=\sqrt{2\cdot6}=\sqrt{12}=\sqrt{4\cdot3}=2\sqrt3\\\\2\sqrt3\cdot\sqrt{18}=2\sqrt{3\cdot18}=2\sqrt{54}=2\sqrt{9\cdot6}=2\cdot3\sqrt6=6\sqrt6\\\\7\sqrt{10}\cdot\frac{1}{2}\sqrt5=7\cdot\frac{1}{2}\cdot\sqrt{10\cdot5}=3,5\cdot\sqrt{50}=3,5\cdot\sqrt{25\cdot2}=3,5\cdot5\sqrt2=17,5\sqrt2[/tex]
Obliczenia od d) do f)
[tex]\sqrt{50}+\sqrt2=\sqrt{25\cdot2}+\sqrt2=5\sqrt2+\sqrt2=6\sqrt2\\\\\sqrt{300}-2\sqrt3=\sqrt{100\cdot3}-2\sqrt3=10\sqrt3-2\sqrt3=8\sqrt3\\\\-2\sqrt{32}-2\sqrt8=-2\sqrt{16\cdot2}-2\sqrt{4\cdot2}=-2\cdot4\sqrt2-2\cdot2\sqrt2=-8\sqrt2-4\sqrt2=-12\sqrt2[/tex]