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Problema (NETO, A.C.M.; T\u00f3picos de Matem\u00e1tica Elementar<\/strong> – N\u00fameros reais, volume 1, SBM Cole\u00e7\u00e3o do Professor de Matem\u00e1tica – Adaptado)<\/span> \n<\/div>\n Lembretes<\/p>\n \n [tex] \\textcolor{#800000}{(1)}[\/tex] Seja [tex]ax^{2} +bx+c=0[\/tex] uma equa\u00e7\u00e3o quadr\u00e1tica, cujas ra\u00edzes s\u00e3o [tex]x_{1}, x_{2}[\/tex] e [tex]a \\neq 0[\/tex]. Ent\u00e3o: \n<\/div>\n Solu\u00e7\u00e3o <\/p>\n Fa\u00e7a [tex]\\boxed{z =7 + 4 \\sqrt{3}}\\,[\/tex] e [tex]\\,\\boxed{ w =7 – 4 \\sqrt{3}}\\,[\/tex] e observe que: Substituindo [tex]\\textcolor{#800000}{(i)}\\;[\/tex] e [tex]\\;\\textcolor{#800000}{(ii)}\\;[\/tex] em [tex]\\;\\textcolor{#800000}{(iii)}[\/tex], obtemos: [tex]\\qquad z^{3} + w^{3} = 14 (z^{2} + w^{2}) – (z+ w) [\/tex] [tex]\\qquad z^{4} + w^{4} = 14 (z^{3} + w^{3}) – (z^{2}+ w^{2})[\/tex] [tex]\\qquad z^{5} + w^{5} = 14 (z^{4} + w^{4}) – (z^{3}+ w^{3})[\/tex] Finalmente, como Solu\u00e7\u00e3o elaborada pelos Moderadores do Blog<\/strong>.<\/p>\n<\/div>\n \n","protected":false},"excerpt":{"rendered":" Problema (Indicado a partir do 9\u00ba ano do E. F.) (NETO, A.C.M.; T\u00f3picos de Matem\u00e1tica Elementar – N\u00fameros reais, volume 1, SBM Cole\u00e7\u00e3o do Professor de Matem\u00e1tica – Adaptado) Encontre o valor de [tex]\\boxed{S= (7 + 4 \\sqrt{3})^{5} + (7 – 4 \\sqrt{3}) ^{5}}[\/tex], sem expandir as pot\u00eancias envolvidas. Lembretes [tex] \\textcolor{#800000}{(1)}[\/tex] Seja [tex]ax^{2} +bx+c=0[\/tex] … <\/p>\n
\n (Indicado a partir do 9\u00ba ano do E. F.)<\/span><\/p>\n
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\nEncontre o valor de [tex]\\boxed{S= (7 + 4 \\sqrt{3})^{5} + (7 – 4 \\sqrt{3}) ^{5}}[\/tex], sem expandir as pot\u00eancias envolvidas.<\/p>\n
\n [tex]\\qquad x_{1} +x_{2} = \\dfrac{-b}{a} [\/tex];
\n[tex] \\qquad x_{1} \\cdot x_{2}= \\dfrac{c}{a}[\/tex].
\n[tex] \\textcolor{#800000}{(2)}[\/tex] [tex]y^{n+m}=y^{n} y^{m}[\/tex], para todo [tex]y[\/tex] real e [tex]n,m[\/tex] n\u00fameros naturais. <\/p>\n
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\n[tex]\\qquad \\qquad z + w = 14\\;[\/tex] e [tex]\\;z \\cdot w = 7^{2} – (4 \\sqrt{3})^{2} = 49-48=1[\/tex].
\nComo [tex] \\textcolor{red}{z+w = 14}\\,[\/tex] e [tex]\\, \\textcolor{blue}{ z \\cdot w =1}[\/tex], segue do Lembrete 1<\/strong>, que [tex]z\\;[\/tex] e [tex]\\; w[\/tex] s\u00e3o ra\u00edzes da equa\u00e7\u00e3o quadr\u00e1tica [tex]x^{2}- \\textcolor{red}{14}x+\\textcolor{blue}{1}=0[\/tex]. Dessa forma,
\n[tex]\\qquad z^{2}= 14z – 1 \\qquad \\;\\;\\;\\, \\textcolor{#800000}{(i)}[\/tex]
\n[tex] \\qquad w^{2}= 14w – 1. \\qquad \\textcolor{#800000}{(ii)}[\/tex]
\nSomando as igualdades [tex]\\textcolor{#800000}{(i)}\\;[\/tex] e [tex]\\; \\textcolor{#800000}{(ii)}[\/tex] obtemos:
\n[tex] \\qquad z^{2} +w^{2} = 14(\\textcolor{red}{z +w}) – 2=14 \\cdot \\textcolor{red}{14} -2= 196 -2 [\/tex]
\n[tex] \\qquad \\boxed{z^{2} +w^{2} = 194} .[\/tex]
\nEm geral, para todo [tex]k[\/tex] natural temos:
\n[tex]\\qquad z^{k+2} + w ^{k+2} =z^{k}z^{2} + w^{k}w^{2}, \\qquad \\textcolor{#800000}{(iii)}[\/tex]
\nde acordo com o Lembrete 2<\/strong>.<\/p>\n
\n[tex]\\qquad z^{k+2} + w^{k+2} = z^{k}(14z-1) + w^{k}(14w-1) [\/tex]
\n[tex]\\qquad z^{k+2} + w^{k+2} =14 (z^{k+1} + w^{k+1}) – (z^{k}+ w^{k}). \\qquad \\textcolor{#800000}{(iv)}[\/tex] <\/p>\n\n
\n[tex]\\qquad z^{3} + w^{3}=14 \\cdot 194-14=2716-14 [\/tex]
\n[tex]\\qquad \\boxed{z^{3} + w^{3}=2702}.[\/tex] <\/p>\n
\n[tex]\\qquad z^{4} + w^{4}= 14 \\cdot 2702-194=37828 -194 [\/tex]
\n[tex]\\qquad \\boxed{z^{4} + w^{4}=37634}[\/tex].<\/p>\n
\n[tex]\\qquad z^{5} + w^{5} =14 \\cdot 37634- 2702[\/tex]
\n[tex]\\qquad z^{5} + w^{5}=524174. \\qquad \\textcolor{#800000}{(v)}[\/tex]<\/ul>\n
\n[tex]\\qquad S= (7 + 4 \\sqrt{3})^{5} + (7 – 4 \\sqrt{3}) ^{5}=z^5+w^5[\/tex],
\nsegue de [tex]\\textcolor{#800000}{(v)}\\;[\/tex] que [tex]\\,\\fcolorbox{black}{#eee0e5}{$S=524174$}\\,.[\/tex]<\/p>\n
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