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Old 2020-10-22, 07:43   #5
LaurV
Romulan Interpreter
 
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Jun 2011
Thailand

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Quote:
Originally Posted by petrw1 View Post
(√√7! + √√7!) * (√(√7! + (√7!)/7!)) = 142
Let \(\alpha=\sqrt{7!}\). You have \(2\sqrt\alpha\cdot\sqrt{\alpha+\frac{\alpha}{\alpha^2}}\). Which, when multiply the radicals and simplify the fraction under it, becomes \(2\sqrt{\alpha^2+1}\). Now substitute back the \(\alpha\), you have \(2\sqrt{7!+1}\), or \(2\sqrt{5041}\), which is 2*71.
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