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Question Number 105018 by bemath last updated on 25/Jul/20

(((√((√5)+2)) + (√((√5)−2)))/(√((√5)+1))) ?

$$\frac{\sqrt{\sqrt{\mathrm{5}}+\mathrm{2}}\:+\:\sqrt{\sqrt{\mathrm{5}}−\mathrm{2}}}{\sqrt{\sqrt{\mathrm{5}}+\mathrm{1}}}\:? \\ $$

Commented by bemath last updated on 25/Jul/20

thank you all

$${thank}\:{you}\:{all} \\ $$

Answered by som(math1967) last updated on 25/Jul/20

let (((√((√5)+2))+(√((√5)−2)))/(√((√5)+1)))=p ⇒{(√((√5)+2))+(√((√5)−2))}^2 =p^2 ((√5)+1) (√5)+2+(√5)−2+2(√(5−2^2 ))=p^2 ((√5)+1{ 2(√5)+2=p^2 ((√5)+1) ∴p^2 =((2((√5)+1))/(((√5)+1)))=2 ∴p=(√2) ans

$$\mathrm{let}\:\frac{\sqrt{\sqrt{\mathrm{5}}+\mathrm{2}}+\sqrt{\sqrt{\mathrm{5}}−\mathrm{2}}}{\sqrt{\sqrt{\mathrm{5}}+\mathrm{1}}}=\mathrm{p} \\ $$$$\Rightarrow\left\{\sqrt{\sqrt{\mathrm{5}}+\mathrm{2}}+\sqrt{\sqrt{\mathrm{5}}−\mathrm{2}}\right\}^{\mathrm{2}} =\mathrm{p}^{\mathrm{2}} \left(\sqrt{\mathrm{5}}+\mathrm{1}\right) \\ $$$$\sqrt{\mathrm{5}}+\mathrm{2}+\sqrt{\mathrm{5}}−\mathrm{2}+\mathrm{2}\sqrt{\mathrm{5}−\mathrm{2}^{\mathrm{2}} }=\mathrm{p}^{\mathrm{2}} \left(\sqrt{\mathrm{5}}+\mathrm{1}\left\{\right.\right. \\ $$$$\mathrm{2}\sqrt{\mathrm{5}}+\mathrm{2}=\mathrm{p}^{\mathrm{2}} \left(\sqrt{\mathrm{5}}+\mathrm{1}\right) \\ $$$$\therefore\mathrm{p}^{\mathrm{2}} =\frac{\mathrm{2}\left(\sqrt{\mathrm{5}}+\mathrm{1}\right)}{\left(\sqrt{\mathrm{5}}+\mathrm{1}\right)}=\mathrm{2} \\ $$$$\therefore\mathrm{p}=\sqrt{\mathrm{2}}\:\mathrm{ans} \\ $$

Answered by Dwaipayan Shikari last updated on 25/Jul/20

a^2 =(((√5)+2+(√5)−2+2(√(5−4)))/((√5)+1))=2(((√5)+1)/((√5)+1))=2 {take it as a} a=(√2)

$${a}^{\mathrm{2}} =\frac{\sqrt{\mathrm{5}}+\mathrm{2}+\sqrt{\mathrm{5}}−\mathrm{2}+\mathrm{2}\sqrt{\mathrm{5}−\mathrm{4}}}{\sqrt{\mathrm{5}}+\mathrm{1}}=\mathrm{2}\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\sqrt{\mathrm{5}}+\mathrm{1}}=\mathrm{2} \\ $$$$\left\{{take}\:{it}\:{as}\:{a}\right\} \\ $$$${a}=\sqrt{\mathrm{2}} \\ $$$$ \\ $$

Answered by john santu last updated on 25/Jul/20

((√(((√((√5)+2))+(√((√5)−2)))^2 ))/(√((√5)+1))) = ((√(((√((√5)+2)))^2 +2(√(((√5)+2)((√5)−2)))+((√((√5)−2)))^2 ))/(√((√5)+1))) = ((√((√5)+2+2(√(5−4))+(√5)−2))/(√((√5)+1))) = ((√(2(√5)+2))/(√((√5)+1))) = (((√2) .(√((√5)+1)))/(√((√5)+1))) = (√2) (JS ⊛)

$$\frac{\sqrt{\left(\sqrt{\sqrt{\mathrm{5}}+\mathrm{2}}+\sqrt{\sqrt{\mathrm{5}}−\mathrm{2}}\right)^{\mathrm{2}} }}{\sqrt{\sqrt{\mathrm{5}}+\mathrm{1}}}\:=\: \\ $$$$\frac{\sqrt{\left(\sqrt{\sqrt{\mathrm{5}}+\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{2}\sqrt{\left(\sqrt{\mathrm{5}}+\mathrm{2}\right)\left(\sqrt{\mathrm{5}}−\mathrm{2}\right)}+\left(\sqrt{\sqrt{\mathrm{5}}−\mathrm{2}}\right)^{\mathrm{2}} }}{\sqrt{\sqrt{\mathrm{5}}+\mathrm{1}}} \\ $$$$=\:\frac{\sqrt{\sqrt{\mathrm{5}}+\mathrm{2}+\mathrm{2}\sqrt{\mathrm{5}−\mathrm{4}}+\sqrt{\mathrm{5}}−\mathrm{2}}}{\sqrt{\sqrt{\mathrm{5}}+\mathrm{1}}} \\ $$$$=\:\frac{\sqrt{\mathrm{2}\sqrt{\mathrm{5}}+\mathrm{2}}}{\sqrt{\sqrt{\mathrm{5}}+\mathrm{1}}}\:=\:\frac{\sqrt{\mathrm{2}}\:.\sqrt{\sqrt{\mathrm{5}}+\mathrm{1}}}{\sqrt{\sqrt{\mathrm{5}}+\mathrm{1}}}\:=\:\sqrt{\mathrm{2}}\: \\ $$$$\left({JS}\:\circledast\right)\: \\ $$

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