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Question-70161




Question Number 70161 by TawaTawa last updated on 01/Oct/19
Answered by mind is power last updated on 01/Oct/19
let a=(√((√5)+2))  b=(√((√5)−2))  a^2 +b^2 =2(√5)  ab=1  ⇒(a+b)=(√(a^2 +b^2 +2ab))=(√(2(√5)+2))=(√2).(√((√5)+1))  ⇒((a+b)/( (√(√(5+1)))))=(√2)  (√(3−2(√2)))=(√(((√2)−1)^2 ))=(√2)−1  answer=1
$${let}\:{a}=\sqrt{\sqrt{\mathrm{5}}+\mathrm{2}}\:\:{b}=\sqrt{\sqrt{\mathrm{5}}−\mathrm{2}} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{2}\sqrt{\mathrm{5}} \\ $$$${ab}=\mathrm{1} \\ $$$$\Rightarrow\left({a}+{b}\right)=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{ab}}=\sqrt{\mathrm{2}\sqrt{\mathrm{5}}+\mathrm{2}}=\sqrt{\mathrm{2}}.\sqrt{\sqrt{\mathrm{5}}+\mathrm{1}} \\ $$$$\Rightarrow\frac{{a}+{b}}{\:\sqrt{\sqrt{\mathrm{5}+\mathrm{1}}}}=\sqrt{\mathrm{2}} \\ $$$$\sqrt{\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}}=\sqrt{\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}} }=\sqrt{\mathrm{2}}−\mathrm{1} \\ $$$${answer}=\mathrm{1} \\ $$$$ \\ $$
Commented by TawaTawa last updated on 01/Oct/19
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by mind is power last updated on 01/Oct/19
y′re Welcom
$${y}'{re}\:{Welcom} \\ $$

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