Question

Proszę o pomoc w rozwiązaniu pierwiastków !!!

Answer 1

Odpowiedź:

1) 7√1/169 = 7*1/13 = 7/13

2) 6√9 + 4∛-125 = 6*3 + 4*(-5) = 18 - 20 = -2

3) [(5∛-216)/9] - 2 = 5*(-6)/9 - 2 = -30/9 - 2 = -3 3/9 - 2 = -1 3/9 = -1 1/3

4) 7√3 - √3 = 6√3

5) √2 + 4∛2 + ∛-2 = √2 + 4∛2 -∛2 = √2 + 3∛2 ≈

1,4142 + 3*1,2599 ≈ 1,4142 + 3,7797 ≈ 5,1939

6) -5∛10 - ∛10 + 6∛10 - 8∛10 = -8∛10

7) 4∛2 - 2,5 - 6,5∛2 + 4 = 1,5 - 2,5∛2

8) 2(5 + 3√6) + 3(√6 - 4) = 10 + 6√6 + 3√6 - 12 = -2 + 9√6

9) 2(3 - ∛3) - (5∛3 + 4) = 6 - 2∛3 - 5∛3 - 4 = 2 - 7∛3

10) -2(5 - ∛7) - 3(2∛7 + 3) = -10 + 2∛7 - 6∛7 - 9 = -19 - 4∛7

-----------

Szczegółowe wyjaśnienie:

Answer 2

[tex]1) \ 7\sqrt{\frac{1}{169}} = 7\frac{\sqrt{1}}{\sqrt{169}} = 7\cdot\frac{1}{13} = \frac{7}{13}\\\\2) \ 6\sqrt{9}+4\sqrt[3]{-125} = 6\cdot3+4\cdot(-5) = 18-20 = -2\\\\3) \ \frac{5\sqrt[3]{-216}}{9}-2 = \frac{5\sqrt[3]{(-6)^{3}}}{9}-2 = \frac{5\cdot(-6)}{9}-2=-\frac{30}{9}-2 = -\frac{10}{3}-2 = -3\frac{1}{3}-2 = -5\frac{1}{3}[/tex]

[tex]4) \ 7\sqrt{3}-\sqrt{3} = 6\sqrt{3}\\\\5) \ \sqrt{2}+4\sqrt[3]{2}+\sqrt[3]{-2} = \sqrt{2}+4\sqrt[3]{2}-\sqrt[3]{2} = \sqrt{2}+3\sqrt[3]{2}[/tex]

[tex]6) \ -5\sqrt[3]{10}-\sqrt[3]{-10}+6\sqrt[3]{10}-8\sqrt[3]{10} = -5\sqrt[3]{10}+\sqrt[3]{10}+6\sqrt[3]{10}-8\sqrt[3]{10} =-6\sqrt[3]{10}\\\\7) \ 4\sqrt[3]{2}-2,5-6,5\sqrt[3]{2}+4 = 1,5 - 2,5\sqrt[3]{2}[/tex]

[tex]8) \ 2(5+3\sqrt{6})+3(\sqrt{6}-4)=2\cdot5+2\cdot3\sqrt{6}+3\cdot\sqrt{6}+3\cdot(-4) = 10+6\sqrt{6}+3\sqrt{6}-12=\\\\=9\sqrt{6}-2[/tex]

[tex]9) \ 2(3-\sqrt[3]{3})-(5\sqrt[3]{3}+4) = 2\cdot3+2\cdot(-\sqrt[3]{3})-5\sqrt[3]{3}-4=6-2\sqrt[3]{3}-5\sqrt[3]{3}-4=2-7\sqrt[3]{3}[/tex]

[tex]10) \ -2(5-\sqrt[3]{7})-3(2\sqrt[3]{7}+3) = -2\cdot5-2\cdot(-\sqrt[3]{7})-3\cdot2\sqrt[3]{7}-3\cdot3=\\\\=-10+2\sqrt[3]{7}-6\sqrt[3]{7}-9=-19-4\sqrt[3]{7}[/tex]