Odpowiedź:
zadanie 1
[tex]2 \sqrt{3} \times \sqrt{10} = 2 \sqrt{30} [/tex]
[tex] \frac{15 \sqrt{6} }{5 \sqrt{2} } = 3 \sqrt{3} [/tex]
[tex]7 \sqrt[3]{4} \times \sqrt[3]{5} = 7 \sqrt[3]{20} [/tex]
[tex] \frac{20 \sqrt[3]{12} }{4 \sqrt[3]{ - 3} } = 5 \sqrt[3]{ - 4} [/tex]
zadanie 2
[tex] \sqrt{28} = \sqrt{4 \times 7} = 2 \sqrt{7} [/tex]
[tex] \sqrt{ {3}^{5} } = \sqrt{243} = \sqrt{81 \times 3} = 9 \sqrt{3} [/tex]
[tex] \sqrt[3]{250} = \sqrt[3]{125 \times 2} = 5 \sqrt[3]{2} [/tex]
zadanie 3
F-trójkąt prostokątny
a=4√3 -przyprostokątna
b=4 -przyprostokątna
c=8 -przeciwprostokątna
[tex]P = \frac{1}{2} \times a \times b \\P = \frac{1}{2} \times 4 \sqrt{3} \times 4 \\ P = 2 \times 4 \sqrt{3} \\P = 8 \sqrt{3} [/tex]
zadanie 4
[tex] \sqrt{48} + 2 \sqrt{3} = 4 \sqrt{3} + 2 \sqrt{3} = 6 \sqrt{3} [/tex]
[tex] \frac{ \sqrt{125} - \sqrt{20} }{ \sqrt{5} } = \frac{5 \sqrt{5} - 2 \sqrt{5} }{ \sqrt{5} } = \frac{3 \sqrt{5} }{ \sqrt{5} } = 3[/tex]
[tex] \sqrt[3]{375} + \sqrt[3]{ - 192} = 5 \sqrt[3]{3} - 4\sqrt[3]{3} = \sqrt[3]{3} [/tex]
