Question

1) sa se calculeze suma:
radical din 8 plus radical din 18 plus radical din 32 - radical sin 200
2)Sa se scoata factorii de sub radical
radical din 2 la a 3 si radical de ordinul 3 din 3 la puterea a 5
radical din -2 la puterea a 6 si radical de ordinul 3 din -2 la puterea a 7

Answer 1

1)√8+√18+√32-√200=
2√2+3√2+4√2-10√2=-√2

2)√2³=2√2
∛3⁵=∛(3³·3²)=3∛3²=3∛9
√(-2)⁶=√2⁶=2³=8

Answer 2

    
[tex]1) \\ \sqrt{8} + \sqrt{18} + \sqrt{32} + \sqrt{200} = \\ =\sqrt{4 \times 2} + \sqrt{9 \times 2} + \sqrt{16 \times 2} - \sqrt{100 \times 2} = \\ =2\sqrt{2} + 3\sqrt{2} + 4\sqrt{2} - 10\sqrt{2} = \\ =\sqrt{2}(2+3+4-10)=\sqrt{2} \times (-1) = \boxed{-\sqrt{2}} \\ \\ 2) \\ \sqrt{2^3} = 2 \sqrt{2} \\ \sqrt[3]{3^5} = \sqrt[3]{3^3 \times 3^2} = \sqrt[3]{3^3} \times \sqrt[3]{3^2}=\boxed{3\sqrt[3]{3^2}} \\ \sqrt{(-2)^6} = \sqrt{2^6} = \sqrt{(2^3)^2}=\boxed{2^3} = \boxed{8} \\ [/tex]

[tex]\sqrt[3]{(-2)^7} =\sqrt[3]{(-2)^6 \times (-2)} =\sqrt[3]{(-2)^6} \times \sqrt[3]{(-2)}= \\ =\sqrt[3]{2^6} \times \sqrt[3]{(-2)}= \sqrt[3]{(2^2)^3} \times \sqrt[3]{(-2)}=\boxed{2^2 \times \sqrt[3]{(-2)}}=\boxed{4 \sqrt[3]{(-2)}}[/tex]