Prove that: $\sqrt{10+\sqrt{24}+\sqrt{40}+\sqrt{60}}=\sqrt{2}+\sqrt3+\sqrt5$
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04.03.2013 21:16
To easy just add in square and win $10+\sqrt{24}+\sqrt{40}+\sqrt{60}=10+\sqrt{24}+\sqrt{40}+\sqrt{60}$
05.07.2013 03:28
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&t=511127&hilit=Equation
14.05.2023 11:49
Squaring both sides of the equation we get: $10+ \sqrt{24} + \sqrt{40} + \sqrt{60} =(\sqrt{2} + \sqrt{3} + \sqrt{5})^2$ $\Rightarrow 10+2 \sqrt{6} +2 \sqrt{10} +2 \sqrt{15} =2+3+5+2(\sqrt{6} + \sqrt{15} + \sqrt{10})$ $ \Rightarrow 10+2 \sqrt{6} +2 \sqrt{10} +2 \sqrt{15} =10+2 \sqrt{6} +2 \sqrt{15} +2 \sqrt{10}$ $\blacksquare$
14.05.2023 11:53
dangerousliri wrote: Prove that: $\sqrt{10+\sqrt{24}+\sqrt{40}+\sqrt{60}}=\sqrt{2}+\sqrt3+\sqrt5$ Let $\sqrt{2} = x, \sqrt{3} =y, \sqrt{5}= z$ $LHS = \sqrt{x^2 + y^2 + z^2 + 2xy + 2yz + 2zx} = x+ y+ z = RHS$