Zadania rozwiazuje z gory na dol od lewego gornego rogu:
[tex]a)\\5\sqrt6+3\sqrt6=8\sqrt6\\b)\\4\sqrt7-2\sqrt7=2\sqrt7\\c)\\5\sqrt[3]5-2\sqrt[3]5+7\sqrt[3]5=3\sqrt[3]5+7\sqrt[3]5=10\sqrt[3]5\\d) \\\sqrt3*\sqrt{12}=\sqrt{3*3*4}=3*2=6\\e)\\\sqrt{25-16}=\sqrt9=3\\f)\\\sqrt{0,064}*\sqrt{1000}=\sqrt{64}=8\\g)\\\frac{\sqrt{450}}{\sqrt2}=\sqrt{\frac{450}2}=\sqrt{225}=15\\h)\\\sqrt[3]{-0,027}-\sqrt[3]{0,064}=-0.3-0,4=-0.7\\i)\\\frac{\sqrt[3]8}{\sqrt[3]{216}}=\sqrt[3]{\frac8{216}}=\sqrt[3]{\frac1{27}}=\frac13\\[/tex]
[tex]j)\\\frac{\sqrt[3]{250}}{\sqrt[3]2}=\sqrt[3]{\frac{250}2}=\sqrt[3]{125}=5\\k) \\2\sqrt[3]5*2\sqrt[3]2=4\sqrt[3]{10}\\l)\\5\sqrt8-2\sqrt{32}=5\sqrt{4*2}-2\sqrt{4*4*2}=5*2\sqrt2-2*4\sqrt2=10\sqrt2-8\sqrt2=2\sqrt2\\m)\\3\sqrt{24}+3\sqrt{150}-\sqrt6=3\sqrt{4*6}+3\sqrt{25*6}-\sqrt6=3*2\sqrt6+3*5\sqrt6-\sqrt6=6\sqrt6+15\sqrt6-\sqrt6=6\sqrt6+14\sqrt6=20\sqrt6\\[/tex]
[tex]n)\\\sqrt[3]{16}+\sqrt[3]{54}-\sqrt[3]{128}=\sqrt[3]{8*2}+\sqrt[3]{27*2}-\sqrt[3]{2*64}=2\sqrt[3]2+3\sqrt[3]2-4\sqrt[3]2=5\sqrt[3]2-4\sqrt[3]2=\sqrt[3]2\\o)\\\sqrt[3]{8+27+29}=\sqrt[3]{64}=4\\p)\\\sqrt[3]{(-3)^3}=-3\\r)\\(\sqrt[3]{-5^3})^2=(-5)^2=25\\s)\\11\sqrt{33}:\sqrt{3}=\frac{11\sqrt{3}*\sqrt{11}}{\sqrt3}=11\sqrt{11}[/tex]
[tex]t)\\\sqrt[3]{-62\frac12*2}=\sqrt[3]{-125}=-5\\u)\\\sqrt2*\sqrt{24\frac12}=\sqrt{49}=7\\w)\\\sqrt{2\sqrt[3]8}=\sqrt{2*2}=2\\x)\\\frac{\sqrt{72}*\sqrt2}{24}=\frac{\sqrt{9*8}*\sqrt2}{24}=\frac{3*2\sqrt2*\sqrt2}{24}=\frac{6*2}{24}=\frac{12}{24}=\frac12\\y)\\\frac{5\sqrt3*3\sqrt5}{15}=\frac{15\sqrt{15}}{15}=\sqrt{15}\\z)\\\frac3{\sqrt[3]{0.27}:\sqrt[3]{-10}}=\frac3{\sqrt[3]{-0.027}}=\frac3{-0.3}=-10[/tex]
