rozłóż liczby na czynniki pierwsze i oblicz wartości pierwiastków:a).... Question from @izkuu01

October 2018 1 30 Report

rozłóż liczby na czynniki pierwsze i oblicz wartości pierwiastków:

a) $\sqrt{324} \sqrt{576} \sqrt{484} \sqrt{1296} \sqrt{4356} \sqrt[3]{216} \sqrt[3]{-729} \sqrt[3]{1728}$

unicorn05 $\sqrt{324}$

324 | 2
162 | 2
   81 | 3
27 | 3
    9 | 3
    3 | 3
   1 | 1

324 = 2²·3⁴

$\sqrt{324}=\sqrt{2^2\cdot3^4}=\sqrt{2^2\cdot3^2\cdot3^2}=2\cdot3\cdot3=18$
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$\sqrt{576}$

576 | 2
288 | 2
144 | 2
72 | 2
36 | 2
   18 | 2
    9 | 3
3 | 3
   1 | 1

576 = 2⁶·3²

$\sqrt{576}=\sqrt{2^6\cdot3^2}=\sqrt{(2^3)^2\cdot3^2}=\sqrt{8^2\cdot3^2}=8\cdot3=24$
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$\sqrt{484}$

484 | 2
242 | 2
121 | 11
11 | 11
   1 | 1

484 = 2²·11²

$\sqrt{484}=\sqrt{2^2\cdot11^2}=2\cdot11=22$
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$\sqrt{1296}$

1296 | 2
648 | 2
324 | 2
   162 | 2
   81 | 3
27 | 3
9 | 3
3 | 3
   1 | 1

1296 = 2⁴·3⁴

$\sqrt{1296}=\sqrt{2^4\cdot3^4}=\sqrt{(2^2)^2\cdot(3^2)^2}=\sqrt{4^2\cdot9^2}=4\cdot9=36$
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$\sqrt{4356}$

4356 | 2
2178 | 2
1089 | 3
363 | 3
121 | 11
    11 | 11
1 | 1

4356 = 2²·3²·11²

$\sqrt{4356}=\sqrt{2^2\cdot3^2\cdot11^2}=2\cdot3\cdot11=66$
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$\sqrt[3]{216}$

216 | 2
108 | 2
54 | 2
27 | 3
    9 | 3
    3 | 3
   1 | 1

216 = 2³·3³

$\sqrt[3]{216} =\sqrt[3]{2^3\cdot3^3}=2\cdot3=6$
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Ponieważ pierwiastek jest nieparzystego stopnia (3), to możemy wyciągnąć "-" przed znak pierwiastka:
$\sqrt[3]{-729}=-\sqrt[3]{729}$

729 | 3
243 | 3
   81 | 3
27 | 3
    9 | 3
    3 | 3
   1 | 1

729 = 3⁶

$\sqrt[3]{-729}=-\sqrt[3]{729}=-\sqrt[3]{3^6}=-\sqrt[3]{(3^2)^3}=-\sqrt[3]{9^3}=-9$
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$\sqrt[3]{1728}$

1728 | 2
864 | 2
432 | 2
   216 | 2
   108 | 2
54 | 2
    27 | 3
      9 | 3
      3 | 3
       1 | 1

1728 = 2⁶·3³

$\sqrt[3]{1728} =\sqrt[3]{2^6\cdot3^3}=\sqrt[3]{2^3\cdot2^3\cdot3^3}=2\cdot2\cdot3=12$

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