Question

Daca A = 20 la a doua - 16 la a doua supra 3 la a doua + 12 la a doua + 9 la a Doua supra 5 la a doua, aratati ca radical din A este numar natural.

Answer 1

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Answer 2

$a = \frac{ {20}^{2} - {16}^{2} }{ {3}^{2} } + \frac{ {12}^{2} + {9}^{2} }{ {5}^{2} } \\ \\ a = \frac{ {(10 \times 2)}^{2} - {2}^{8} }{9} + \frac{ {(4 \times 3)}^{2} + {(3 \times 3)}^{2} }{ {5}^{2} } \\ \\ a = \frac{ {10}^{2} \times {2}^{2} - {2}^{8} }{9} + \frac{ {4}^{2} \times {3}^{2} + {3}^{2} \times {3}^{2} }{ {5}^{2} } \\ \\ a = \frac{ {(10}^{2} - {2}^{6}) \times {2}^{2} }{9} + \frac{ {( {4} }^{2} + {3}^{2}) \times {3}^{2} }{ {5}^{2} } \\ \\ a = \frac{(100 - 64) \times 4}{9} + \frac{(16 + 9) \times {3}^{2} }{ {5}^{2} } \\ \\ a = \frac{36 \times 4}{9} + \frac{25 \times {3}^{2} }{ {5}^{2} } \\ \\ a = \frac{144}{9} + \frac{ {5}^{2} \times {3}^{2} }{ {5}^{2} } \\ \\ a = 16 + {3}^{2} \\ \\ a = 16 + 9 \\ \\ a = 25 \\ \\ \sqrt{a} = \sqrt{25} = \sqrt{ {5}^{2} } = 5$