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Problema Calcule o valor de [tex]4x+2y[\/tex] sabendo que [tex]\\sqrt{x+\\sqrt{y}}+\\sqrt{x-\\sqrt{y}}=2[\/tex] e [tex]\\sqrt{y+\\sqrt{x}}+\\sqrt{y-\\sqrt{x}}=1[\/tex].<\/p>\n<\/div>\n \n<\/div>\n \n Solu\u00e7\u00e3o 1<\/p>\n Considerando [tex]x,y\\ge0[\/tex] reais, ao elevar ambos os lados das equa\u00e7\u00f5es ao quadrado, obtemos: Isolando ambas as ra\u00edzes das equa\u00e7\u00f5es e elevando novamente ao quadrado, temos: Da\u00ed, temos um sistema de equa\u00e7\u00f5es bastante simples, onde, ap\u00f3s somar as equa\u00e7\u00f5es, obtemos [tex]y=\\dfrac {5}{3}[\/tex], e da\u00ed [tex]x=\\dfrac{17}{12}[\/tex].<\/p>\n Por fim, basta fazermos: <\/p>\n [tex]\\qquad 4x+2y=4\\cdot\\bigg(\\dfrac{17}{12}\\bigg)+2\\cdot \\bigg(\\dfrac{5}{3}\\bigg)=\\dfrac{27}{3}=9.[\/tex]<\/p>\n Solu\u00e7\u00e3o elaborada pelo COM Phidias. <\/p>\n Solu\u00e7\u00e3o 2<\/p>\n Primeiramente, vamos construir um sistema com as equa\u00e7\u00f5es fornecidas: Elevando ambas as equa\u00e7\u00f5es ao quadrado, temos o seguinte: Como [tex]\\sqrt{a} \\cdot \\sqrt{b} = \\sqrt{ab}[\/tex], e [tex](a+b) \\cdot (a-b)=a^2-b^2[\/tex], podemos substituir os termos seguintes: Assim, ficamos com: Donde obtemos: Novamente, elevando ambos os termos ao quadrado, nas duas equa\u00e7\u00f5es, ficamos com: Por fim, ficamos com o seguinte sistema: Agora, utilizando o m\u00e9todo de substitui\u00e7\u00e3o, vem: Por fim, podemos apenas substituir seus valores, e obter o resultado de [tex]4x+2y[\/tex]: Assim, conclu\u00edmos que o valor de [tex]4x+2y[\/tex] \u00e9 igual a [tex]9[\/tex].<\/p>\n Solu\u00e7\u00e3o elaborada pelo COM Pot\u00eancias de Euler. <\/p>\n \n <\/p>\n","protected":false},"excerpt":{"rendered":" Problema (Indicado a partir do 3\u00ba ano do E. M.) Calcule o valor de [tex]4x+2y[\/tex] sabendo que [tex]\\sqrt{x+\\sqrt{y}}+\\sqrt{x-\\sqrt{y}}=2[\/tex] e [tex]\\sqrt{y+\\sqrt{x}}+\\sqrt{y-\\sqrt{x}}=1[\/tex]. Solu\u00e7\u00e3o 1 Considerando [tex]x,y\\ge0[\/tex] reais, ao elevar ambos os lados das equa\u00e7\u00f5es ao quadrado, obtemos: [tex]\\qquad \\large x+\\sqrt{y}+2\\bigg(\\sqrt{x+\\sqrt{y}}\\bigg)\\bigg(\\sqrt{x-\\sqrt{y}}\\bigg)+x-\\sqrt{y}=2^2 \\large \\implies 2x+2\\sqrt{x^2-y}=4[\/tex] [tex]\\qquad\\large y+\\sqrt{x}+2\\bigg(\\sqrt{y+\\sqrt{x}}\\bigg)\\bigg(\\sqrt{y-\\sqrt{x}}\\bigg)+y-\\sqrt{x}=1^2 \\large \\implies 2y+2\\sqrt{y^2-x}=1.[\/tex] Isolando ambas as ra\u00edzes das equa\u00e7\u00f5es e elevando novamente … <\/p>\n
\n (Indicado a partir do 3\u00ba ano do E. M.)<\/span><\/p>\n
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\n[tex]\\qquad \\large x+\\sqrt{y}+2\\bigg(\\sqrt{x+\\sqrt{y}}\\bigg)\\bigg(\\sqrt{x-\\sqrt{y}}\\bigg)+x-\\sqrt{y}=2^2 \\large \\implies 2x+2\\sqrt{x^2-y}=4[\/tex]
\n[tex]\\qquad\\large y+\\sqrt{x}+2\\bigg(\\sqrt{y+\\sqrt{x}}\\bigg)\\bigg(\\sqrt{y-\\sqrt{x}}\\bigg)+y-\\sqrt{x}=1^2 \\large \\implies 2y+2\\sqrt{y^2-x}=1.[\/tex]<\/p>\n
\n[tex]\\qquad\\large 2\\sqrt{x^2-y}=4-2x \\implies 4x^2-4y=16-16x+4x^2 \\implies 4x-y=4[\/tex]
\n[tex]\\qquad
\n\\large 2\\sqrt{y^2-x}=1-2y \\implies 4y^2-4x=1-4y+4y^2 \\implies 4y-4x=1.
\n[\/tex]<\/p>\n
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\n[tex]\\qquad\\begin{cases}
\n\\sqrt{x+\\sqrt{y}}+\\sqrt{x-\\sqrt{y}}=2 \\\\
\n\\sqrt{y+\\sqrt{x}}+\\sqrt{y-\\sqrt{x}}=1.
\n\\end{cases}[\/tex]<\/p>\n
\n[tex]\\qquad\\begin{cases}
\n(\\sqrt{x+\\sqrt{y}})^2+2(\\sqrt{x+\\sqrt{y}} \\cdot \\sqrt{x-\\sqrt{y}})+(\\sqrt{x-\\sqrt{y}})^2=4 \\\\
\n(\\sqrt{y+\\sqrt{x}})^2+2(\\sqrt{y+\\sqrt{x}} \\cdot \\sqrt{y-\\sqrt{x}})+(\\sqrt{y-\\sqrt{x}})^2=1.
\n\\end{cases}[\/tex]<\/p>\n
\n[tex]\\qquad\\sqrt{x+\\sqrt{y}} \\cdot \\sqrt{x-\\sqrt{y}}=\\sqrt{(x+\\sqrt{y}) \\cdot (x-\\sqrt{y})}=\\sqrt{x^2-y}\\\\
\n\\qquad \\sqrt{y+\\sqrt{x}} \\cdot \\sqrt{y-\\sqrt{x}}=\\sqrt{(y+\\sqrt{x}) \\cdot (y-\\sqrt{x})}=\\sqrt{y^2-x}.[\/tex]<\/p>\n
\n[tex]\\qquad\\begin{cases}
\n(\\sqrt{x+\\sqrt{y}})^2+2(\\sqrt{x^2-y})+(\\sqrt{x-\\sqrt{y}})^2=4 \\Rightarrow x+\\sqrt{y}+2(\\sqrt{x^2-y})+x-\\sqrt{y}=4 \\\\
\n(\\sqrt{y+\\sqrt{x}})^2+2(\\sqrt{y^2-x})+(\\sqrt{y-\\sqrt{x}})^2=1 \\Rightarrow y+\\sqrt{x}+2(\\sqrt{y^2-x})+y-\\sqrt{x}=1.
\n\\end{cases}[\/tex]<\/p>\n
\n[tex]\\qquad\\begin{cases}
\n2x+2(\\sqrt{x^2-y})=4 \\Rightarrow 2(x+\\sqrt{x^2-y})=4 \\Rightarrow x+\\sqrt{x^2-y}=2 \\Rightarrow \\sqrt{x^2-y}=2-x\\\\
\n2y+2(\\sqrt{y^2-x})=1 \\Rightarrow 2(y+\\sqrt{y^2-x})=1 \\Rightarrow y+\\sqrt{y^2-x}=\\frac{1}{2} \\Rightarrow \\sqrt{y^2-x}=\\frac{1}{2}-y.
\n\\end{cases}[\/tex]<\/p>\n
\n[tex]\\qquad\\begin{cases}
\nx^2-y=(2-x)^2 \\Rightarrow x^2-y=4-4x+x^2 \\Rightarrow -y=4-4x\\\\
\ny^2-x=\\left(\\frac{1}{2}-y\\right)^2 \\Rightarrow y^2-x=\\frac{1}{4}-y+y^2 \\Rightarrow -x=\\frac{1}{4}-y.
\n\\end{cases}[\/tex]<\/p>\n
\n[tex]\\qquad\\begin{cases}
\ny=4x-4 \\\\
\nx=y-\\frac{1}{4}.
\n\\end{cases}[\/tex]<\/p>\n
\n[tex]\\qquad x=(4x-4)-\\frac{1}{4} \\Rightarrow 3x=\\frac{1}{4}+4 \\Rightarrow 3x=\\frac{1}{4}+\\frac{16}{4} \\Rightarrow 3x=\\frac{17}{4} \\therefore x=\\frac{17}{12} \\\\
\n\\qquad y=4 \\cdot \\frac{17}{12}-4 \\Rightarrow y=\\frac{17}{3}-\\frac{12}{3} \\therefore y=\\frac{5}{3}.
\n[\/tex]<\/p>\n
\n[tex]\\qquad4x+2y=4 \\cdot \\frac{17}{12}+2 \\cdot \\frac{5}{3}=\\frac{68}{12}+\\frac{10}{3}=\\frac{68}{12}+\\frac{40}{12}=\\frac{108}{12}=9.[\/tex]<\/p>\n
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