Szczegółowe wyjaśnienie:
Zad.5
[tex]a)\ \sqrt3\cdot\sqrt5\cdot\sqrt{15}=\sqrt{3\cdot5\cdot15}=\sqrt{15\cdot15}=\sqrt{15^2}=15\\\\\\b)\ \sqrt{\dfrac{4}{5}}\cdot\qrt{0,6}\cdot\sqrt{12}=\sqrt{\dfrac{4\!\!\!\!\diagup^2}{5}\cdot\dfrac{6}{10\!\!\!\!\!\diagup_5}\cdot12}=\sqrt{\dfrac{2\cdot6\cdot12}{5\cdot5}}=\sqrt{\dfrac{12\cdot12}{5^2}}\\\\=\sqrt{\dfrac{12^2}{5^2}}=\sqrt{\left(\dfrac{12}{5}\right)^2}=\dfrac{12}{5}=2,4[/tex]
[tex]c)\ \sqrt[3]{1\dfrac{1}{5}}\cdot\sqrt[3]{2,5}\cdot\sqrt[3]{1\dfrac{1}{8}}=\sqrt[3]{\dfrac{6\!\!\!\!\diagup^3}{5\!\!\!\!\diagup_1}\cdot\dfrac{25\!\!\!\!\!\diagup^{5\!\!\!\!\diagup^1}}{10\!\!\!\!\!\diagup_{2\!\!\!\!\diagup_1}}\cdot\dfrac{9}{8}}=\sqrt[3]{\dfrac{3\cdot9}{8}}=\sqrt[3]{\dfrac{27}{8}}=\dfrac{\sqrt[3]{27}}{\sqrt[3]8}=\dfrac{3}{2}=1,5[/tex]
skorzystałem ze wzorów:
[tex]\sqrt{a\cdot b}=\sqrt{a}\cdot\sqrt{b},\ a,b\geq0\\\\\sqrt[3]{a\cdot b}=\sqrt[3]{a}\cdot\sqrt[3]{b}\\\\\sqrt{a^2}=a,\ a\geq0\\\\\sqrt[3]{\dfrac{a}{b}}=\dfrac{\sqrt[3]{a}}{\sqrt[3]{b}},\ b\neq0[/tex]
Zad.10
[tex]a)\ a=3\sqrt2cm,\ b=\sqrt2cm\\\\P=a\cdot b\to P=3\sqrt2\cdot\sqrt2=3\cdot2=6(cm^2)\\\\L=2a+2b\to L=2\cdot3\sqrt2+2\cdot\sqrt2=6\sqrt2+2\sqrt2=8\sqrt2(cm)\\\\\\b)\ a=\sqrt{12}cm,\ b=\sqrt3cm\\\\P=a\cdot b\to P=\sqrt{12}\cdot\sqrt3=\sqrt{12\cdot3}=\sqrt{36}=6(cm^2)\\\\L=2a+2b\to L=2\cdot\sqrt{12}+2\cdot\sqrt3=2\sqrt{4\cdot3}+2\sqrt3=2\cdot\sqrt4\cdot\sqrt3+2\sqrt3\\\\=2\cdot2\cdot\sqrt3+2\sqrt3=4\sqrt2+2\sqrt3=6\sqrt3(cm)[/tex]
