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Question-200568

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Question Number 200568 by Rupesh123 last updated on 20/Nov/23
Commented by AST last updated on 20/Nov/23
Cannot be uniquely determined.
$${Cannot}\:{be}\:{uniquely}\:{determined}. \\ $$
Commented by Frix last updated on 20/Nov/23
Let α=∠cd 0<f<∞ ⇔ 0<α<((2π)/5)
$$\mathrm{Let}\:\alpha=\angle{cd} \\ $$$$\mathrm{0}<{f}<\infty\:\Leftrightarrow\:\mathrm{0}<\alpha<\frac{\mathrm{2}\pi}{\mathrm{5}} \\ $$
Commented by Rupesh123 last updated on 20/Nov/23
Nice one, sir!
Answered by a.lgnaoui last updated on 20/Nov/23
△ABC BC^2 =AB^2 +AC^2 ( e+d)^2 =(((1+(√5) )^2 )/4)+(c+f)^2 ⇒ (d+e)^2 −(c+f)^2 =((3+(√5))/2) (1) △CDE CD^2 =DE^2 +CE^2 c^2 −d^2 =(1/4) (2) △ABC / CDE Semblables: ((AC)/(BC))=((EC)/(DC)) ((c+f)/(d+e)) =(d/c) ⇒ (((c+f)^2 )/((d+e)^2 ))=(d^2 /c^2 )=(d^2 /(d^2 +(1/4)))(3) ((1+(√5))/((c+f)))=(1/d) d=((c+f)/(1+(√5))) (4) ⇒c+f=(1+(√5) )d (d+e)^2 =(1+(√5) )^2 d^2 +((3+(√5))/2) (2) (1+(1/(4d^2 )))(c+f)^2 −(1+(√5) )^2 d^2 =((3+(√5))/2) ⇒(c+f)^2 = ((4(6+2(√5) )d^4 +(6+2(√5) )d^2 )/(4d^2 +1)) f^2 +(((1+(√5))/2))^2 =e^2 +(1/4) e^2 −f^2 =((5+2(√5))/4) f=(√(e^2 −((5+2(√5))/4))) (5) (1)⇒(d+e)^2 =(c+(√(e^2 −((5+2(√5))/4))) )^2 +((3+(√5))/2) d^2 +e^2 +2de=c^2 +e^2 −((5+2(√5))/4)+((3+(√5))/2) +2c((√(e^2 −((5+2(√5))/4))) ) [ 2de−c(√(4e^2 −5−2(√5) )) =(1/2) 4de=(√((4d^2 +1)(4e^2 −5−2(√5) ))) =1 (i) (d+e)=(√(2(3+(√5))d^2 +((3+(√5))/2))) (ii) 2d+2e=2(√((1/2)(3+(√5) )×(2d)^2 +((3+(√5))/2))) 2d=x 2e=y (ii) x+y=2(√((1/2)(3+(√5) )(1+x^2 ))) (i) xy =(√((x^2 +1)(y^2 −5−2(√5) )) )−1 ⇒(x+y)^2 = 2(3+(√5) )(1+x^2 ) (xy+1)^2 =( x^2 +1)(y^2 −5−2(√5) ) { ((x^2 +y^2 +2xy=2(3+(√5) )x^2 +2(3+(√5) ))),((x^2 y^2 +2xy+1=x^2 y^2 −(5+2(√5) )x^2 )) :} +y^2 −(5+2(√5) (5+2(√5) )x^2 +y^2 +2xy=2(3+(√5) ) (I) 2xy =y^2 −(5+2(√5) )x^2 −(6+2(√5) ) (II) (I)−(II)⇒ (5+2(√5) )x^2 +y^2 =2(3+(√5) )−y^2 + (5+2(√5) )x^2 +(6+2(√5) ) ⇒ y=(√(6+2(√5))) alors e=((√(2(3+(√5) )))/2) (ii) x=2(√((1/2)(3+(√5) )(1+x^2 ))) −(√(6+2(√5))) ⇒ 2(3+(√5) )(1+x^2 )=(x+(√(6+2(√5)[)) )^2 (6+2(√5) )x^2 +6+2(√5) =x^2 +6+2(√5) + 2x(√(6+2(√5))) (5+2(√5) )x^2 −2x(√(6+2(√5) )) =0 x=((2(√(6+2(√5))))/(5+2(√5))) ⇒ d=((√(6+2(√5)))/(5+2(√5))) { ((d=((√(6+2(√5)))/(5+2(√5))))),((e=((√(6+2(√5)))/2))) :} calcul de f (5)⇒ f=(√(e^2 −((5+2(√5))/4))) =(√(((6+2(√5))/4)−((5+2(√5))/4))) ⇒ f=(1/2)
$$\bigtriangleup\boldsymbol{\mathrm{ABC}}\:\:\:\:\boldsymbol{\mathrm{BC}}^{\mathrm{2}} =\boldsymbol{\mathrm{AB}}^{\mathrm{2}} +\boldsymbol{\mathrm{AC}}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\left(\:\boldsymbol{\mathrm{e}}+\boldsymbol{\mathrm{d}}\right)^{\mathrm{2}} =\frac{\left(\mathrm{1}+\sqrt{\mathrm{5}}\:\right)^{\mathrm{2}} }{\mathrm{4}}+\left(\boldsymbol{\mathrm{c}}+\boldsymbol{\mathrm{f}}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\Rightarrow\:\:\left(\boldsymbol{\mathrm{d}}+\boldsymbol{\mathrm{e}}\right)^{\mathrm{2}} −\left(\boldsymbol{\mathrm{c}}+\boldsymbol{\mathrm{f}}\right)^{\mathrm{2}} =\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:\:\:\:\left(\mathrm{1}\right) \\ $$$$\bigtriangleup\boldsymbol{\mathrm{CDE}}\:\:\:\:\:\:\boldsymbol{\mathrm{CD}}^{\mathrm{2}} =\boldsymbol{\mathrm{DE}}^{\mathrm{2}} +\boldsymbol{\mathrm{CE}}^{\mathrm{2}} \\ $$$$\:\:\:\:\boldsymbol{\mathrm{c}}^{\mathrm{2}} −\boldsymbol{\mathrm{d}}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}}\:\:\:\:\left(\mathrm{2}\right) \\ $$$$\bigtriangleup\boldsymbol{\mathrm{ABC}}\:/\:\boldsymbol{\mathrm{CDE}}\:\:\boldsymbol{\mathrm{Semblables}}:\:\frac{\boldsymbol{\mathrm{AC}}}{\boldsymbol{\mathrm{BC}}}=\frac{\boldsymbol{\mathrm{EC}}}{\boldsymbol{\mathrm{DC}}} \\ $$$$\:\:\:\frac{\boldsymbol{\mathrm{c}}+\boldsymbol{\mathrm{f}}}{\boldsymbol{\mathrm{d}}+\boldsymbol{\mathrm{e}}}\:\:\:=\frac{\boldsymbol{\mathrm{d}}}{\boldsymbol{\mathrm{c}}}\:\:\Rightarrow\:\frac{\left(\boldsymbol{\mathrm{c}}+\boldsymbol{\mathrm{f}}\right)^{\mathrm{2}} }{\left(\boldsymbol{\mathrm{d}}+\boldsymbol{\mathrm{e}}\right)^{\mathrm{2}} }=\frac{\mathrm{d}^{\mathrm{2}} }{\mathrm{c}^{\mathrm{2}} }=\frac{\mathrm{d}^{\mathrm{2}} }{\mathrm{d}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}}\left(\mathrm{3}\right) \\ $$$$\:\: \\ $$$$\:\:\:\:\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\left(\boldsymbol{\mathrm{c}}+\boldsymbol{\mathrm{f}}\right)}=\frac{\mathrm{1}}{\boldsymbol{\mathrm{d}}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{d}}=\frac{\boldsymbol{\mathrm{c}}+\boldsymbol{\mathrm{f}}}{\mathrm{1}+\sqrt{\mathrm{5}}}\:\:\:\:\left(\mathrm{4}\right) \\ $$$$\:\:\Rightarrow\boldsymbol{\mathrm{c}}+\boldsymbol{\mathrm{f}}=\left(\mathrm{1}+\sqrt{\mathrm{5}}\:\right)\boldsymbol{\mathrm{d}}\: \\ $$$$\:\:\:\left(\boldsymbol{\mathrm{d}}+\boldsymbol{\mathrm{e}}\right)^{\mathrm{2}} =\left(\mathrm{1}+\sqrt{\mathrm{5}}\:\right)^{\mathrm{2}} \boldsymbol{\mathrm{d}}^{\mathrm{2}} +\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:\:\:\left(\mathrm{2}\right) \\ $$$$\:\: \\ $$$$\:\:\:\:\:\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}\boldsymbol{\mathrm{d}}^{\mathrm{2}} }\right)\left(\boldsymbol{\mathrm{c}}+\boldsymbol{\mathrm{f}}\right)^{\mathrm{2}} −\left(\mathrm{1}+\sqrt{\mathrm{5}}\:\right)^{\mathrm{2}} \boldsymbol{\mathrm{d}}^{\mathrm{2}} =\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$ \\ $$$$\Rightarrow\left(\mathrm{c}+\mathrm{f}\right)^{\mathrm{2}} =\:\frac{\mathrm{4}\left(\mathrm{6}+\mathrm{2}\sqrt{\mathrm{5}}\:\right)\boldsymbol{\mathrm{d}}^{\mathrm{4}} +\left(\mathrm{6}+\mathrm{2}\sqrt{\mathrm{5}}\:\right)\boldsymbol{\mathrm{d}}^{\mathrm{2}} }{\mathrm{4}\boldsymbol{\mathrm{d}}^{\mathrm{2}} +\mathrm{1}} \\ $$$$ \\ $$$$\boldsymbol{\mathrm{f}}^{\mathrm{2}} +\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{2}} =\boldsymbol{\mathrm{e}}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\:\:\:\boldsymbol{\mathrm{e}}^{\mathrm{2}} −\boldsymbol{\mathrm{f}}^{\mathrm{2}} =\frac{\mathrm{5}+\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{4}} \\ $$$$ \\ $$$$\:\:\:\boldsymbol{\mathrm{f}}=\sqrt{\boldsymbol{\mathrm{e}}^{\mathrm{2}} −\frac{\mathrm{5}+\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{4}}}\:\:\:\:\:\:\left(\mathrm{5}\right)\: \\ $$$$ \\ $$$$\:\:\left(\mathrm{1}\right)\Rightarrow\left(\boldsymbol{\mathrm{d}}+\boldsymbol{\mathrm{e}}\right)^{\mathrm{2}} =\left(\boldsymbol{\mathrm{c}}+\sqrt{\boldsymbol{\mathrm{e}}^{\mathrm{2}} −\frac{\mathrm{5}+\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{4}}}\:\right)^{\mathrm{2}} +\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{d}}^{\mathrm{2}} +\boldsymbol{\mathrm{e}}^{\mathrm{2}} +\mathrm{2}\boldsymbol{\mathrm{de}}=\boldsymbol{\mathrm{c}}^{\mathrm{2}} +\boldsymbol{\mathrm{e}}^{\mathrm{2}} −\frac{\mathrm{5}+\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{4}}+\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\:+\mathrm{2}\boldsymbol{\mathrm{c}}\left(\sqrt{\boldsymbol{\mathrm{e}}^{\mathrm{2}} −\frac{\mathrm{5}+\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{4}}}\:\right) \\ $$$$ \\ $$$$\:\:\:\left[\:\:\mathrm{2}\boldsymbol{\mathrm{de}}−\boldsymbol{\mathrm{c}}\sqrt{\mathrm{4}\boldsymbol{\mathrm{e}}^{\mathrm{2}} −\mathrm{5}−\mathrm{2}\sqrt{\mathrm{5}}\:}\:=\frac{\mathrm{1}}{\mathrm{2}}\right. \\ $$$$ \\ $$$$\:\:\:\mathrm{4}\boldsymbol{\mathrm{de}}=\sqrt{\left(\mathrm{4}\boldsymbol{\mathrm{d}}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{4}\boldsymbol{\mathrm{e}}^{\mathrm{2}} −\mathrm{5}−\mathrm{2}\sqrt{\mathrm{5}}\:\right)}\:=\mathrm{1}\:\:\left(\boldsymbol{\mathrm{i}}\right) \\ $$$$\:\:\:\: \\ $$$$\:\:\left(\boldsymbol{\mathrm{d}}+\boldsymbol{\mathrm{e}}\right)=\sqrt{\mathrm{2}\left(\mathrm{3}+\sqrt{\mathrm{5}}\right)\boldsymbol{\mathrm{d}}^{\mathrm{2}} +\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}}}\:\:\:\:\left(\boldsymbol{\mathrm{ii}}\right) \\ $$$$ \\ $$$$ \\ $$$$\:\:\:\mathrm{2}\boldsymbol{\mathrm{d}}+\mathrm{2}\boldsymbol{\mathrm{e}}=\mathrm{2}\sqrt{\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{3}+\sqrt{\mathrm{5}}\:\right)×\left(\mathrm{2}\boldsymbol{\mathrm{d}}\right)^{\mathrm{2}} +\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}}} \\ $$$$\:\:\:\:\mathrm{2}\boldsymbol{\mathrm{d}}=\boldsymbol{\mathrm{x}}\:\:\:\:\:\:\mathrm{2}\boldsymbol{\mathrm{e}}=\boldsymbol{\mathrm{y}} \\ $$$$ \\ $$$$\:\:\left(\boldsymbol{\mathrm{ii}}\right)\:\:\:\:\:\:\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{y}}=\mathrm{2}\sqrt{\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{3}+\sqrt{\mathrm{5}}\:\right)\left(\mathrm{1}+\boldsymbol{\mathrm{x}}^{\mathrm{2}} \right)} \\ $$$$\left.\:\:\left(\boldsymbol{\mathrm{i}}\right)\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{xy}}\:\:\:\:=\sqrt{\left(\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\mathrm{1}\right)\left(\boldsymbol{\mathrm{y}}^{\mathrm{2}} −\mathrm{5}−\mathrm{2}\sqrt{\mathrm{5}}\:\right.}\:\right)−\mathrm{1} \\ $$$$ \\ $$$$\:\:\:\:\:\Rightarrow\left(\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{y}}\right)^{\mathrm{2}} =\:\:\mathrm{2}\left(\mathrm{3}+\sqrt{\mathrm{5}}\:\right)\left(\mathrm{1}+\boldsymbol{\mathrm{x}}^{\mathrm{2}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\left(\boldsymbol{\mathrm{xy}}+\mathrm{1}\right)^{\mathrm{2}} =\left(\:\:\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\mathrm{1}\right)\left(\boldsymbol{\mathrm{y}}^{\mathrm{2}} −\mathrm{5}−\mathrm{2}\sqrt{\mathrm{5}}\:\right) \\ $$$$\:\:\:\:\begin{cases}{\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{y}}^{\mathrm{2}} +\mathrm{2}\boldsymbol{\mathrm{xy}}=\mathrm{2}\left(\mathrm{3}+\sqrt{\mathrm{5}}\:\right)\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\mathrm{2}\left(\mathrm{3}+\sqrt{\mathrm{5}}\:\right)}\\{\boldsymbol{\mathrm{x}}^{\mathrm{2}} \boldsymbol{\mathrm{y}}^{\mathrm{2}} +\mathrm{2}\boldsymbol{\mathrm{xy}}+\mathrm{1}=\boldsymbol{\mathrm{x}}^{\mathrm{2}} \boldsymbol{\mathrm{y}}^{\mathrm{2}} −\left(\mathrm{5}+\mathrm{2}\sqrt{\mathrm{5}}\:\right)\boldsymbol{\mathrm{x}}^{\mathrm{2}} }\end{cases} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\boldsymbol{\mathrm{y}}^{\mathrm{2}} −\left(\mathrm{5}+\mathrm{2}\sqrt{\mathrm{5}}\right. \\ $$$$ \\ $$$$ \\ $$$$\:\:\left(\mathrm{5}+\mathrm{2}\sqrt{\mathrm{5}}\:\right)\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{y}}^{\mathrm{2}} +\mathrm{2}\boldsymbol{\mathrm{xy}}=\mathrm{2}\left(\mathrm{3}+\sqrt{\mathrm{5}}\:\right)\:\:\:\:\:\:\left(\boldsymbol{\mathrm{I}}\right) \\ $$$$\:\:\mathrm{2}\boldsymbol{\mathrm{xy}}\:\:=\boldsymbol{\mathrm{y}}^{\mathrm{2}} −\left(\mathrm{5}+\mathrm{2}\sqrt{\mathrm{5}}\:\right)\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\left(\mathrm{6}+\mathrm{2}\sqrt{\mathrm{5}}\:\right)\:\:\:\left(\boldsymbol{\mathrm{II}}\right) \\ $$$$ \\ $$$$\left(\boldsymbol{\mathrm{I}}\right)−\left(\boldsymbol{\mathrm{II}}\right)\Rightarrow \\ $$$$\:\:\:\left(\mathrm{5}+\mathrm{2}\sqrt{\mathrm{5}}\:\right)\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{y}}^{\mathrm{2}} =\mathrm{2}\left(\mathrm{3}+\sqrt{\mathrm{5}}\:\right)−\boldsymbol{\mathrm{y}}^{\mathrm{2}} + \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{5}+\mathrm{2}\sqrt{\mathrm{5}}\:\right)\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\left(\mathrm{6}+\mathrm{2}\sqrt{\mathrm{5}}\:\right) \\ $$$$ \\ $$$$\Rightarrow\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{y}}=\sqrt{\mathrm{6}+\mathrm{2}\sqrt{\mathrm{5}}} \\ $$$$\boldsymbol{\mathrm{alors}}\:\:\:\:\:\:\:\boldsymbol{\mathrm{e}}=\frac{\sqrt{\mathrm{2}\left(\mathrm{3}+\sqrt{\mathrm{5}}\:\right)}}{\mathrm{2}} \\ $$$$\:\:\left(\boldsymbol{\mathrm{ii}}\right)\:\:\:\boldsymbol{\mathrm{x}}=\mathrm{2}\sqrt{\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{3}+\sqrt{\mathrm{5}}\:\right)\left(\mathrm{1}+\boldsymbol{\mathrm{x}}^{\mathrm{2}} \right)}\:−\sqrt{\mathrm{6}+\mathrm{2}\sqrt{\mathrm{5}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\Rightarrow\:\mathrm{2}\left(\mathrm{3}+\sqrt{\mathrm{5}}\:\right)\left(\mathrm{1}+\boldsymbol{\mathrm{x}}^{\mathrm{2}} \right)=\left(\boldsymbol{\mathrm{x}}+\sqrt{\mathrm{6}+\mathrm{2}\sqrt{\mathrm{5}}\left[\right.}\:\right)^{\mathrm{2}} \\ $$$$\left(\mathrm{6}+\mathrm{2}\sqrt{\mathrm{5}}\:\right)\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\mathrm{6}+\mathrm{2}\sqrt{\mathrm{5}}\:=\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\mathrm{6}+\mathrm{2}\sqrt{\mathrm{5}}\:+ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\boldsymbol{\mathrm{x}}\sqrt{\mathrm{6}+\mathrm{2}\sqrt{\mathrm{5}}} \\ $$$$\left(\mathrm{5}+\mathrm{2}\sqrt{\mathrm{5}}\:\right)\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\mathrm{2}\boldsymbol{\mathrm{x}}\sqrt{\mathrm{6}+\mathrm{2}\sqrt{\mathrm{5}}\:}\:=\mathrm{0} \\ $$$$\:\:\:\boldsymbol{\mathrm{x}}=\frac{\mathrm{2}\sqrt{\mathrm{6}+\mathrm{2}\sqrt{\mathrm{5}}}}{\mathrm{5}+\mathrm{2}\sqrt{\mathrm{5}}}\:\:\:\:\:\:\:\Rightarrow\:\boldsymbol{\mathrm{d}}=\frac{\sqrt{\mathrm{6}+\mathrm{2}\sqrt{\mathrm{5}}}}{\mathrm{5}+\mathrm{2}\sqrt{\mathrm{5}}} \\ $$$$\:\:\begin{cases}{\boldsymbol{\mathrm{d}}=\frac{\sqrt{\mathrm{6}+\mathrm{2}\sqrt{\mathrm{5}}}}{\mathrm{5}+\mathrm{2}\sqrt{\mathrm{5}}}}\\{\boldsymbol{\mathrm{e}}=\frac{\sqrt{\mathrm{6}+\mathrm{2}\sqrt{\mathrm{5}}}}{\mathrm{2}}}\end{cases} \\ $$$$\: \\ $$$$\boldsymbol{\mathrm{calcul}}\:\boldsymbol{\mathrm{de}}\:\boldsymbol{\mathrm{f}}\:\: \\ $$$$\left(\mathrm{5}\right)\Rightarrow\:\:\:\boldsymbol{\mathrm{f}}=\sqrt{\boldsymbol{\mathrm{e}}^{\mathrm{2}} −\frac{\mathrm{5}+\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{4}}} \\ $$$$\:\:\:\:=\sqrt{\frac{\mathrm{6}+\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{4}}−\frac{\mathrm{5}+\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{4}}}\: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\Rightarrow\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{f}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$
Commented by a.lgnaoui last updated on 20/Nov/23

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