Question

Calculați, după ce ați scos factorii de sub radical:
a)√160-( √810- √9610+ √6760)
b) √4365-( √3087+ √4732- √2268)

Answer 1

$a) \sqrt{160} - ( \sqrt{810} - \sqrt{9610} + \sqrt{6760}) = \\ \\ \sqrt{160} = \sqrt{2 \times 2 \times 2 \times 2 \times 2 \times 5} = \sqrt{(2 \times 2) \times (2 \times 2) \times 2 \times 5} = \sqrt{ {2}^{2} \times {2}^{2} \times 2 \times 5 } = (2 \times 2) \sqrt{2 \times 5} = 4 \sqrt{10} \\ \\ \sqrt{810} = \sqrt{2 \times 3 \times 3 \times 3 \times 3 \times 5} = \sqrt{(3 \times 3) \times (3 \times 3) \times 2 \times 5} = \sqrt{ {3}^{2} \times {3}^{2} \times 2 \times 5 } = (3 \times 3) \sqrt{2 \times 5} = 9 \sqrt{10} \\ \\ \sqrt{9610} = \sqrt{2 \times 5 \times 31 \times 31} = \sqrt{(31 \times 31) \times 2 \times 5} = \sqrt{ {31}^{2} \times 2 \times 5 } = 31 \sqrt{2 \times 5} = 31 \sqrt{10} \\ \\ \sqrt{6760} = \sqrt{2 \times 2 \times 2 \times 5 \times 13 \times 13} = \sqrt{(2 \times 2) \times (13 \times 13) \times 2 \times 5} = \sqrt{ {2}^{2} \times {13}^{2} \times 2 \times 5 } = (2 \times 13) \sqrt{2 \times 5} = 26 \sqrt{10} \\ \\ 4 \sqrt{10} - (9 \sqrt{10} - 31 \sqrt{10} + 26 \sqrt{10}) = \\ \\ 4 \sqrt{10} - (9 - 31 + 26) \sqrt{10} = \\ \\ 4 \sqrt{10} - (35 - 31) \sqrt{10} = \\ \\ 4 \sqrt{10} - 4 \sqrt{10} = \\ \\ 0$

$b) \sqrt{4365} - ( \sqrt{3087} + \sqrt{4732} - \sqrt{2268}) = \\ \\ \sqrt{4365} = \sqrt{3 \times 3 \times 5 \times 97} = \sqrt{(3 \times 3) \times 5 \times 97} = \sqrt{ {3}^{2} \times 5 \times 97 } = 3 \sqrt{5 \times 97} = 3 \sqrt{485} \\ \\ \sqrt{3087} = \sqrt{3 \times 3 \times 7 \times 7 \times 7} = \sqrt{(3 \times 3) \times (7 \times 7) \times 7} = \sqrt{ {3}^{2} \times {7}^{2} \times 7 } = (3 \times 7) \sqrt{7} = 21 \sqrt{7} \\ \\ \sqrt{4732} = \sqrt{2 \times 2 \times 7 \times 13 \times 13} = \sqrt{(2 \times 2) \times (13 \times 13) \times 7} = \sqrt{ {2}^{2} \times {13}^{2} \times 7 } = (2 \times 13) \sqrt{7} = 26 \sqrt{7} \\ \\ \sqrt{2268} = \sqrt{2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 7} = \sqrt{(2 \times 2) \times (3 \times 3) \times (3 \times 3) \times 7} = \sqrt{ {2}^{2} \times {3}^{2} \times {3}^{2} \times 7 } = (2 \times 3 \times 3) \sqrt{7} = 18 \sqrt{7} \\ \\ 3 \sqrt{485} - (21 \sqrt{7} + 26 \sqrt{7} - 18 \sqrt{7}) = \\ \\ 3 \sqrt{485} - (21 + 26 - 18) \sqrt{7} = \\ \\ 3 \sqrt{485} - (47 - 18) \sqrt{7} = \\ \\ 3 \sqrt{485} - 29 \sqrt{7}$