Question

Encontre o valor de x na igualdade:​

Answer 1

Resposta:

x=1

Explicação passo-a-passo:

Temos que:

raiz((raiz(5)+1)/(raiz(5)-1)) + raiz((raiz(5)-1)/(raiz(5)+1)) =

(5x)/(25^(1/4))

raiz((raiz(5)+1)/(raiz(5)-1) . (raiz(5)+1)/(raiz(5)+1)) + raiz((raiz(5)-1)/(raiz(5)+1) . (raiz(5)+1)/(raiz(5)+1)) =

(5x)/(25^(1/4))

raiz( (raiz(5)+1)^2 / (raiz(5)^2 -1^2) ) + raiz( (raiz(5)^2 -1^2) / (raiz(5)+1)^2 ) =

(5x)/(25^(1/4))

raiz((raiz(5)+1)^2)/raiz(raiz(5)^2 -1^2) + raiz(raiz(5)^2 -1^2)/raiz((raiz(5)+1)^2) =

(5x)/(25^(1/4))

(raiz(5)+1)/raiz(5-1) + raiz(5-1)/(raiz(5)+1) = (5x)/(25^(1/4))

(raiz(5)+1)/raiz(4) + raiz(4)/(raiz(5)+1) = (5x)/(25^(1/4))

(raiz(5)+1)/2 + 2/(raiz(5)+1) = (5x)/((5^2)^(1/4))

(raiz(5)+1)/2 + 2/(raiz(5)+1) = (5x)/(5^(2/4))

(raiz(5)+1)/2 + 2/(raiz(5)+1) = (5x)/(5^(1/2))

(raiz(5)+1)/2 + 2/(raiz(5)+1) = (5x)/raiz(5) . raiz(5)/raiz(5)

(raiz(5)+1)/2 + 2/(raiz(5)+1) = (5x).raiz(5)/5

(raiz(5)+1)/2 + 2/(raiz(5)+1) = x.raiz(5)

x= (raiz(5)+1)/(2. raiz(5)) + 2/((raiz(5)+1).raiz(5))

x= raiz(5)/(2. raiz(5)) + 1/(2. raiz(5)) + 2/(raiz(5)^2 + raiz(5))

x= 1/2 + (1/2).(1/raiz(5)) + 2/(5 + raiz(5))

x= 1/2 + (1/2).(1/raiz(5)).(raiz(5)/raiz(5)) + 2/(5 + raiz(5))

x= 1/2 + (1/2).(raiz(5)/5) + 2/(5 + raiz(5))

x= 1/2 + raiz(5)/10 + 2/(5 + raiz(5))

x= (5+raiz(5))/10 + 2/(5+raiz(5))

x= [(5+raiz(5))^2 + 20] / 10.(5+raiz(5))

x= [25 + 10.raiz(5) + 5 + 20] / 10.(5+raiz(5))

x= [50 + 10.raiz(5)] / 10.(5+raiz(5))

x= 10.(5+raiz(5)) / 10.(5+raiz(5))

x=1

Blz?

Abs :)

Answer 2

Explicação passo-a-passo:

Racionalizar as raízes

$\sqrt{{(\sqrt{5}+1)(\sqrt{5} +1)\over(\sqrt{5} -1)(\sqrt{5} +1)} } +\sqrt{{{(\sqrt{5}-1)(\sqrt{5} -1)\over(\sqrt{5}+1)(\sqrt{5}  -1)} } }={5x\over\sqrt[4]{5^2} }\\\\  \\{ \sqrt{(\sqrt{5}+1)^2\over(\sqrt{5})^2-1^2}  } +\sqrt{{(\sqrt{5} -1)^2\over(\sqrt{5} )^2-1^2}} ={5x\over\sqrt{2} }\\ \\ \\ \sqrt{{(\sqrt{5} +1)^2\over5-1}} +\sqrt{{(\sqrt{5}-1)^2\over5-1}}={5x\over\sqrt{2}  } \\ \\ \\ \sqrt{{(\sqrt{5} +1)^2\over4}} +\sqrt{{(\sqrt{5}-1)^2\over4} } ={5x\over\sqrt{2} }$

${\sqrt{5} +1\over2}+{\sqrt{5} -1\over2}={5x\over\sqrt{2} }\\\\  \\ {\sqrt{5} +1+\sqrt{5} -1\over2}={5x\over\sqrt{2} }\\ \\ \\ {\not2\sqrt{5} \over\not2}={5x\over\sqrt{5} }\\ \\ \sqrt{5} ={5x\over\sqrt{5} }\\ \\ 5x=\sqrt{5} .\sqrt{5} \\ \\ 5x=\sqrt{25} \\ \\ 5x=5\\ \\ x={5\over5}\\ \\\fbox{ x=1 }$