Odpowiedź i szczegółowe wyjaśnienie:
ZADANIE 1.
[tex]\sqrt[3]{0,125}=\sqrt[3]{(0,5)^3}=0,5\\\\\sqrt{2500}=\sqrt{50^2}=50\\\\\sqrt{\dfrac{36}{49}}=\sqrt{\dfrac{6^2}{7^2}}=\sqrt{(\dfrac67)^2}=\dfrac67\\\\\\\sqrt{2\dfrac79}=\sqrt{\dfrac{25}{9}}=\sqrt{\dfrac{5^2}{3^2}}=\sqrt{(\dfrac53)^2}=\dfrac53\\\\\sqrt[3]{1728}=\sqrt[3]{12^3}=12\\\\\sqrt{3\dfrac{6}{25}}=\sqrt{\dfrac{81}{25}}=\sqrt{\dfrac{9^2}{5^2}}=\sqrt{(\dfrac95)^2}=\dfrac95\\\\\\\sqrt[3]{\dfrac1{27}}=\sqrt[3]{(\dfrac{1}{3})^3}=\dfrac13\\\\\sqrt{10000}=\sqrt{100^2}=100[/tex]
ZADANIE 2.
[tex](\sqrt{1\dfrac79}+\sqrt{1\dfrac{13}{36}})\cdot\sqrt{1\dfrac{17}{64}}=(\sqrt{\dfrac{16}{9}}+\sqrt{\dfrac{49}{36}})\cdot\sqrt{\dfrac{81}{64}}=(\dfrac43+\dfrac{7}{6})\cdot\dfrac{9}{8}=\\\\\\=(\dfrac{8}{6}+\dfrac76)\cdot\dfrac98=\dfrac{15}{6}\cdot\dfrac{9}{8}=\dfrac{15\cdot3}{2\cdot8}=\dfrac{45}{16}[/tex]
ZADANIE 3.
Pole kwadratu wyraża się wzorem:
[tex]P=a^2\\\\P=0,25\\\\a^2=0,25\\\\a=\sqrt{0,25}\\\\a=\sqrt{\dfrac14}=\sqrt{(\dfrac12)^2}=\dfrac12\ [cm][/tex]
Krawędź ma miarę 0,5cm.
Obwód tego kwadratu wynosi:
[tex]Obw=4a\\\\Obw=4\cdot \dfrac12=2\ [cm][/tex]