$\sqrt{53+20\sqrt{7}}$ can be written in the form $a+b\sqrt{c}$, where $a,$ $b,$ and $c$ are integers and $c$ has no factors which is a perf

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$\sqrt{53+20\sqrt{7}}$ can be written in the form $a+b\sqrt{c}$, where $a,$ $b,$ and $c$ are integers and $c$ has no factors which is a perf

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$\sqrt{53+20\sqrt{7}}$ can be written in the form , where and are integers and has no factors which is a perfect square of any positive integer other than 1. Find.$abc$

dududsu2 Nov 18, 2020

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Guest · Answer 1 · 2020-11-18T02:17:10

Simplify the following:
sqrt(20 sqrt(7) + 53)

53 + 20 sqrt(7) = 25 + 20 sqrt(7) + 28 = 25 + 20 sqrt(7) + 4 (sqrt(7))^2 = (2 sqrt(7) + 5)^2:
sqrt((2 sqrt(7) + 5)^2)

Cancel exponents. sqrt((5 + 2 sqrt(7))^2) = 2 sqrt(7) + 5:

5 + 2 sqrt(7) abc =5 x 2 x 7 = 70

dududsu2 · Answer 2 · 2020-11-18T03:40:44

I meant a+b+c

$\sqrt{53+20\sqrt{7}}$ can be written in the form $a+b\sqrt{c}$, where $a,$ $b,$ and $c$ are integers and $c$ has no factors which is a perf

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$\sqrt{53+20\sqrt{7}}$ can be written in the form $a+b\sqrt{c}$, where $a,$ $b,$ and $c$ are integers and $c$ has no factors which is a perf

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