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Question Number 137590 by Ñï= last updated on 04/Apr/21
x=1+((π+(π+1)^2 +(π+2)^3 +(π+3)^4 )/(4+5^2 +6^3 +7^4 ))  (√(x+2(√(x−1))))+(√(x−2(√(x−1))))=?
$${x}=\mathrm{1}+\frac{\pi+\left(\pi+\mathrm{1}\right)^{\mathrm{2}} +\left(\pi+\mathrm{2}\right)^{\mathrm{3}} +\left(\pi+\mathrm{3}\right)^{\mathrm{4}} }{\mathrm{4}+\mathrm{5}^{\mathrm{2}} +\mathrm{6}^{\mathrm{3}} +\mathrm{7}^{\mathrm{4}} } \\ $$$$\sqrt{{x}+\mathrm{2}\sqrt{{x}−\mathrm{1}}}+\sqrt{{x}−\mathrm{2}\sqrt{{x}−\mathrm{1}}}=? \\ $$
Commented by mindispower last updated on 04/Apr/21
(√(x+2(√(x−1))))=(√(((√(x−1))+1)^2 ))=1+(√(x−1))  (√(x−2(√(x−1))))=(√(((√(x−1))−1)^2 ))1−(√(x−1))  1<x<2  we  get 2
$$\sqrt{{x}+\mathrm{2}\sqrt{{x}−\mathrm{1}}}=\sqrt{\left(\sqrt{{x}−\mathrm{1}}+\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{1}+\sqrt{{x}−\mathrm{1}} \\ $$$$\sqrt{{x}−\mathrm{2}\sqrt{{x}−\mathrm{1}}}=\sqrt{\left(\sqrt{{x}−\mathrm{1}}−\mathrm{1}\right)^{\mathrm{2}} }\mathrm{1}−\sqrt{{x}−\mathrm{1}} \\ $$$$\mathrm{1}<{x}<\mathrm{2} \\ $$$${we}\:\:{get}\:\mathrm{2} \\ $$

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