Odpowiedź i szczegółowe wyjaśnienie:
[tex]a)\ \sqrt{4\cdot81}=\sqrt4\cdot\sqrt{81}=2\cdot9=18\\\\b)\ \sqrt2\cdot\sqrt8=\sqrt{2\cdot8}=\sqrt{16}=4\\\\c)\ 3\cdot2\sqrt7=6\sqrt7\\\\d)\ \dfrac{\sqrt{54}}{\sqrt6}=\sqrt{\dfrac{54}{6}}=\sqrt9=3\\\\e)\ 3\sqrt2-5\sqrt2=-2\sqrt2\\\\f)\ 3\sqrt2\cdot4\sqrt2=12\cdot\sqrt4=12\cdot2=24\\\\g)\ 4\sqrt2+\sqrt8=4\sqrt2+2\sqrt2=6\sqrt2\\\\h)\ \sqrt{200}-\sqrt{50}=\sqrt{100\cdot2}-\sqrt{25\cdot2}=10\sqrt2-5\sqrt2=5\sqrt2\\\\[/tex]
[tex]i)\ \sqrt{800}+\sqrt{242}=\sqrt{400\cdot2}+\sqrt{121\cdot2}=20\sqrt2+11\sqrt2=31\sqrt2\\\\j)\ \sqrt{1\dfrac{9}{16}}+\sqrt{2\dfrac{1}{4}}=\sqrt{\dfrac{25}{16}}+\sqrt{\dfrac{9}{4}}=\dfrac{\sqrt{25}}{\sqrt{16}}+\dfrac{\sqrt9}{\sqrt4}=\dfrac{5}{4}+\dfrac32=\dfrac53+\dfrac64=\dfrac{11}{4}=2\dfrac34[/tex]
