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Identidades Alg\u00e9bricas <\/p>\n
Solu\u00e7\u00f5es dos problemas propostos – Primeira Sala<\/p>\n
\nPara ajudar no entendimento das solu\u00e7\u00f5es, relacionamos algumas identidades apresentadas no nosso Blog. <\/p>\n
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[tex]1.[\/tex] Prove que se [tex]a+b+c=0[\/tex], ent\u00e3o:<\/p>\n
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(a)<\/strong> [tex] \\, \\dfrac{a^5+b^5+c^5}{5}=\\dfrac{a^3+b^3+c^3}{3}\\times \\dfrac{a^2+b^2+c^2}{2},[\/tex] \n \n \n [tex]2.[\/tex] (OCM<\/strong>) Se [tex]x^2+x+1=0[\/tex], calcule o valor num\u00e9rico de <\/p>\n \n [tex]\\qquad \\qquad \\left(x+\\dfrac{1}{x}\\right)^2+\\left(x^2+\\dfrac{1}{x^2}\\right)^2+\\left(x^3+\\dfrac{1}{x^3}\\right)^2+\\cdots+\\left(x^{27}+\\dfrac{1}{x^{27}}\\right)^2.[\/tex]<\/p>\n \n Inicialmente, observe que [tex]\\boxed{x^2=-1-x} \\, [\/tex]. [tex]\\quad \\left(x+\\dfrac{1}{x}\\right)^2=\\left(\\dfrac{x^2+1}{x}\\right)^2=\\left(\\dfrac{-x}{x}\\right)^2=1;[\/tex] \n \n [tex]3.[\/tex] (Torneio das Cidades<\/strong>) Sejam [tex]a[\/tex], [tex]b[\/tex], [tex]c[\/tex] e [tex]d[\/tex] n\u00fameros reais tais que<\/p>\n \n [tex]\\qquad \\quad \\qquad a^3+b^3+c^3+d^3=a+b+c+d=0.[\/tex]<\/p>\n \n Prove que a soma de dois desses n\u00fameros \u00e9 zero.<\/p>\n \n \n \n [tex]4.[\/tex] (Col\u00f4mbia<\/strong>) Sejam [tex]a[\/tex], [tex]b[\/tex], [tex]c[\/tex] reais tais que <\/p>\n \n [tex]\\qquad a^{12}+b^{12}+c^{12}=8\\quad [\/tex] e [tex]\\quad \\dfrac{(a-b)^2+(b-c)^2+(c-a)^2}{abc}=-\\dfrac{6}{a+b+c}.[\/tex]<\/p>\n \n Calcule [tex] \\, \\, a^6+b^6+c^6[\/tex].<\/p>\n \n De [tex]\\left(x^2+y^2+z^2 \\right)^2=2(x^4+y^4+z^4), \\, [\/tex] como [tex]a^3=x, \\, b^3=y, \\, c^3=z \\, \\, [\/tex] e [tex] \\, \\, a^{12}+b^{12}+c^{12}=8, \\, [\/tex] segue que: \n \n [tex]5.[\/tex] (China<\/strong>) Seja [tex]\\alpha[\/tex] um n\u00famero real tal que [tex]\\alpha^3-\\alpha-1=0[\/tex]. Determine o valor num\u00e9rico de<\/p>\n \n [tex]\\qquad \\qquad\\sqrt[3]{3\\alpha^2 -4\\alpha}+\\alpha \\sqrt[4]{2\\alpha^2 +3\\alpha +2}.[\/tex]<\/p>\n \n Finalmente, \n \n [tex]6.[\/tex] (OMG<\/strong>) Seja [tex]a=2008[\/tex]. Determine o valor da soma <\/p>\n \n [tex]\\qquad \\qquad \\displaystyle\\sum_{k=-2007}^{k=2007}\\dfrac{1}{1+a^k}.[\/tex]<\/p>\n \n \n assim, adicionando dois termos equidistantes dos extremos, bem como os extremos, obteremos sempre somas iguais a [tex]1[\/tex].<\/li>\n Deste modo, j\u00e1 podemos calcular a soma total: \n \n [tex]7.[\/tex] (REOIM<\/strong>) Demonstrar que \n Como [tex]4>2 \\, [\/tex] e [tex] \\, 9>6[\/tex] s\u00e3o desigualdades trivialmente verdadeiras, segue que as desigualdades abaixo tamb\u00e9m o s\u00e3o: \n \n [tex]8.[\/tex] (Canad\u00e1<\/strong>) Calcule: \n (b)<\/strong> [tex]\\sqrt{1+\\dfrac{2}{5}}\\cdot\\sqrt{1+\\dfrac{2}{6}}\\cdot\\sqrt{1+\\dfrac{2}{7}}\\cdot\\sqrt{1+\\dfrac{2}{8}}\\cdots \\sqrt{1+\\dfrac{2}{57}}\\cdot\\sqrt{1+\\dfrac{2}{58}}.[\/tex]<\/p>\n \n \n \n (b)<\/strong> \n \n [tex]9.[\/tex] (Rom\u00eania<\/strong>) Sejam [tex]x \\, [\/tex] e [tex] \\, y[\/tex] n\u00fameros naturais tais que \n Determinar o valor de [tex]x+y[\/tex].<\/p>\n \n Mas [tex]2^x\\cdot3^y=24^k[\/tex] pode ser reescrito como sendo [tex]2^x\\cdot3^y=(2^3\\cdot3^1)^{15\\cdot59}=2^{3\\cdot 15\\cdot59}\\cdot3^{1\\cdot 15\\cdot59}[\/tex], donde [tex]x=45\\cdot59 \\, [\/tex] e [tex] \\, y=15\\cdot59.[\/tex] \n \n [tex]10.[\/tex] (R\u00fassia<\/strong>) Sejam [tex]a[\/tex] e [tex]b[\/tex] n\u00fameros reais n\u00e3o nulos que satisfazem \u00e0 igualdade \n \n \n [tex]11.[\/tex] (REOIM<\/strong>) Dados os n\u00fameros \n \n \n [tex]12.[\/tex] (OCM<\/strong>) Determine qual \u00e9 o maior desses dois n\u00fameros:<\/p>\n \n [tex]\\qquad \\qquad \\dfrac{123456+10^{999}}{123457+10^{999}} \\qquad [\/tex] ou [tex]\\qquad \\dfrac{123457+10^{999}}{123458+10^{999}}[\/tex]<\/p>\n \n \n Observe que, agora, queremos comparar [tex]\\dfrac{a-1}{a}[\/tex] com [tex]\\dfrac{a}{a+1}[\/tex] e o que sabemos da “lei da tricotomia” \u00e9 apenas que:<\/p>\n e somente uma dessa op\u00e7\u00f5es. [tex]\\qquad \\dfrac{a-1}{a} \\, \\boxed{?} \\, \\dfrac{a}{a+1} \\iff (a-1)(a+1) \\, \\boxed{?} \\, a^2 \\iff a^2-1 \\, \\boxed{?} \\, a^2.[\/tex] \n
\n(b)<\/strong> [tex] \\, \\dfrac{a^7+b^7+c^7}{7}=\\dfrac{a^5+b^5+c^5}{5}\\times \\dfrac{a^2+b^2+c^2}{2}[\/tex].<\/p>\n
\n[tex]\\quad a^2+b^2+c^2=(a+b+c)^2-2ab-2ac-2bc \\\\
\n\\quad a^2+b^2+c^2=-2ab-2ac-2bc \\\\
\n\\quad\\dfrac{a^2+b^2+c^2}{2}=-ab-ac-bc=-ab-c(a+b) \\\\
\n\\quad \\dfrac{a^2+b^2+c^2}{2}=-ab+(a+b)^2=a^2+ab+b^2. \\qquad \\color{#800000}{(I)}[\/tex]
\nVeja tamb\u00e9m que
\n[tex](a+b+c)(a^2+b^2+c^2-ab-ac-bc)=\\\\
\n=a^3+ab^2+ac^2-a^2b-a^2c-abc+ba^2+b^3+bc^2-ab^2-abc-b^2c+ca^2+cb^2+c^3-abc-ac^2-bc^2 =\\\\
\n=a^3+b^3+c^3-3abc \\, [\/tex]
\nou seja,
\n[tex]a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc). [\/tex]
\nAssim, sendo [tex]a+b+c=0, \\, [\/tex] segue que
\n[tex]\\quad a^3+b^3+c^3-3abc=0 [\/tex]
\n[tex]\\quad a^3+b^3+c^3=3abc [\/tex]
\n[tex]\\quad \\dfrac{a^3+b^3+c^3}{3}=abc=-ab(a+b). \\qquad \\color{#800000}{(II)}[\/tex]
\nPor outro lado, [tex](a+b)^5=a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5, \\, [\/tex] assim:
\n[tex]\\quad -c^5=(a+b)^5=a^5+b^5+5(a^4b+2a^3b^2+2a^2b^3+ab^4)[\/tex]
\n[tex]\\quad \\dfrac{a^5+b^5+c^5}{5}=-ab(a^3+2a^2b+2ab^2+b^3)[\/tex]
\n[tex]\\quad \\dfrac{a^5+b^5+c^5}{5}=-ab(a^3+(a^2b+a^2b)+(ab^2+ab^2)+b^3)[\/tex]
\n[tex]\\quad \\dfrac{a^5+b^5+c^5}{5}=-ab((a^3+a^2b)+(a^2b+ab^2)+(ab^2+b^3))[\/tex]
\n[tex]\\quad \\dfrac{a^5+b^5+c^5}{5}=-ab[a^2(a+b)+ab(a+b)+b^2(a+b)][\/tex]
\n[tex]\\quad \\dfrac{a^5+b^5+c^5}{5}=-ab(a+b)(a^2+ab+b^2). \\qquad \\color{#800000}{(III)}[\/tex]
\nFinalmente, observe que, multiplicando as igualdades [tex]\\color{#800000}{(I)}[\/tex] e [tex]\\color{#800000}{(II)}[\/tex], obtemos
\n[tex]\\quad \\begin{align*} \\dfrac{a^2+b^2+c^2}{2} \\times \\dfrac{a^3+b^3+c^3}{3}&=(a^2+ab+b^2) \\times (-ab(a+b))\\\\
\n&=-ab(a+b)(a^2+ab+b^2)\\\\
\n&\\stackrel{\\color{#800000}{(III)}}{=}\\dfrac{a^5+b^5+c^5}{5}.\\end{align*}
\n[\/tex]<\/p>\n
\n
\n(b)<\/strong> Comecemos por
\n[tex]\\quad (a+b)^7=a^7+7a^6b+21a^5b^2+35a^4b^3+35a^3b^4+21a^2b^5+7ab^6+b^7[\/tex]
\n[tex]\\quad -c^7=a^7+b^7+7[a^6b+3a^5b^2+5a^4b^3+5a^3b^4+3a^2b^5+ab^6][\/tex]
\n[tex]\\quad \\dfrac{a^7+b^7+c^7}{7}=-ab[a^5+3a^4b+5a^3b^2+5a^2b^3+3ab^4+b^5][\/tex]
\n[tex]\\quad \\dfrac{a^7+b^7+c^7}{7}=-ab[a^4(a+b)+2a^3b(a+b)+3a^2b^2(a+b)+2ab^3(a+b)+b^4(a+b)][\/tex]
\n[tex]\\quad \\dfrac{a^7+b^7+c^7}{7}=-ab(a+b)(a^4+2a^3b+3a^2b^2+2ab^3+b^4). \\qquad \\color{#800000}{(I)}[\/tex]
\nAgora, veja que
\n[tex]\\quad \\dfrac{a^2+b^2+c^2}{2}=a^2+ab+b^2; \\qquad \\color{#800000}{(II)}[\/tex]
\ne que
\n[tex]\\quad \\dfrac{a^5+b^5+c^5}{5}=-ab(a+b)(a^2+ab+b^2). \\qquad \\color{#800000}{(III)}[\/tex]
\nPortanto, multiplicando as igualdades [tex]\\color{#800000}{(III)}[\/tex] e [tex]\\color{#800000}{(II)}[\/tex], temos:
\n[tex]\\quad \\dfrac{a^5+b^5+c^5}{5} \\times \\dfrac{a^2+b^2+c^2}{2}=-ab(a+b)(a^2+ab+b^2)(a^2+ab+b^2)[\/tex]
\n[tex]\\quad \\dfrac{a^5+b^5+c^5}{5} \\times \\dfrac{a^2+b^2+c^2}{2}=-ab(a+b)(a^2+ab+b^2)^2[\/tex]
\n[tex]\\quad \\dfrac{a^5+b^5+c^5}{5} \\times \\dfrac{a^2+b^2+c^2}{2}=-ab(a+b)(a^4+b^4+2a^3b+3a^2b^2+2ab^3)[\/tex]
\n[tex]\\quad \\dfrac{a^5+b^5+c^5}{5} \\times \\dfrac{a^2+b^2+c^2}{2}=\\dfrac{a^7+b^7+c^7}{7}.[\/tex]\n<\/div>\n<\/div>\n<\/div>\n
\n
\nAssim,
\n[tex]\\qquad x^4=1+2x+x^2=1+2x-1-x=x \\, [\/tex],
\ne
\n[tex]\\qquad x^3=x \\cdot x^2=-x-x^2=-x+1+x=1.[\/tex]
\nDesta forma, podemos escrever:<\/p>\n
\n[tex]\\quad \\left(x^2+\\dfrac{1}{x^2}\\right)^2=\\left(\\dfrac{x^4+1}{x^2}\\right)^2=\\left(\\dfrac{x+1}{-1-x}\\right)^2=1;[\/tex]
\n[tex]\\quad \\left(x^3+\\dfrac{1}{x^3}\\right)^2=(1+1)^2=4;[\/tex]
\n[tex]\\quad \\left(x^4+\\dfrac{1}{x^4}\\right)^2=\\left(x+\\dfrac{1}{x}\\right)^2=1.[\/tex]
\nDa mesma forma, [tex]x^5=x^4\\cdot x=x^2 \\, \\, [\/tex] e, portanto, [tex]\\left(x^5+\\dfrac{1}{x^5}\\right)^2=\\left(x^2+\\dfrac{1}{x^2}\\right)^2=1;[\/tex]
\nE ainda, [tex]x^6=x^5\\cdot x=x^3 \\, \\, [\/tex] e, portanto, [tex]\\left(x^6+\\dfrac{1}{x^6}\\right)^2=\\left(x^3+\\dfrac{1}{x^3}\\right)^2=4.[\/tex]
\nPerceba que h\u00e1 uma periodicidade a cada tr\u00eas pot\u00eancias, de forma que a soma
\n[tex]\\qquad \\qquad \\left(x+\\dfrac{1}{x}\\right)^2+\\left(x^2+\\dfrac{1}{x^2}\\right)^2+\\left(x^3+\\dfrac{1}{x^3}\\right)^2+\\cdots+\\left(x^{27}+\\dfrac{1}{x^{27}}\\right)^2[\/tex]
\npode ser calculada como sendo [tex] \\, \\fcolorbox{black}{#eee0e5}{$9\\cdot(1+1+4)=54$} \\, .[\/tex]\n<\/p><\/div>\n<\/div>\n<\/div>\n
\n
\nAssim, segue que:
\n[tex]\\quad d^3=-(a+b+c)^3=-[a^3+b^3+c^3+3(a+b)(a+c)(b+c)].[\/tex]
\nMas, da hip\u00f3tese [tex]a^3+b^3+c^3+d^3=0 \\, [\/tex], tamb\u00e9m segue que [tex]d^3=-a^3-b^3-c^3 \\, [\/tex] logo,
\n[tex]\\quad -[a^3+b^3+c^3+3(a+b)(a+c)(b+c)]=-a^3-b^3-c^3[\/tex]
\n[tex]\\quad -a^3-b^3-c^3-3(a+b)(a+c)(b+c)=-a^3-b^3-c^3[\/tex]
\n[tex]\\quad -3(a+b)(a+c)(b+c)=0[\/tex]
\n[tex]\\quad \\boxed{(a+b)(a+c)(b+c)=0}.[\/tex]
\nDa \u00faltima igualdade, conclu\u00edmos que, necessariamente, [tex]\\boxed{a+b=0} \\, [\/tex] ou [tex] \\, \\boxed{a+c=0} \\, [\/tex] ou [tex] \\, \\boxed{b+c=0} \\, .[\/tex]\n<\/div>\n<\/div>\n<\/div>\n
\n
\n[tex]\\quad (2a^2+2b^2+2c^2-2ab-2ac-2bc)(a+b+c)=-6abc[\/tex]
\nou, ainda,
\n[tex]\\quad (a^2+b^2+c^2-ab-ac-bc)(a+b+c)=-3abc.[\/tex][tex]\\qquad \\color{#800000}{(I)}[\/tex]
\nMas,
\n[tex](a^2+b^2+c^2-ab-ac-bc)(a+b+c)=\\\\
\n= a^3+ab^2+ac^2-a^2b-a^2c-abc+a^2b+b^3+bc^2-ab^2-abc-b^2c+a^2c+b^2c+c^3-abc-ac^2-bc^2=\\\\
\n=a^3+b^3+c^3-3abc,[\/tex]
\nportanto, de [tex]\\color{#800000}{(I)}[\/tex], segue que
\n[tex]\\quad a^3+b^3+c^3-3abc=-3abc[\/tex]
\ne, ent\u00e3o [tex]\\boxed{a^3+b^3+c^3=0}.[\/tex][tex]\\qquad \\color{#800000}{(II)}[\/tex]
\nPara facilitar o entendimento do que se segue, chame [tex]\\boxed{a^3=x, \\, b^3=y, \\, c^3=z} \\, \\, [\/tex] e, assim, podemos reescrever a igualdade [tex]\\color{#800000}{(II)}[\/tex] como [tex]x+y+z=0.[\/tex]
\nAgora, observe que, se [tex]x+y+z=0, [\/tex] ent\u00e3o [tex]\\left(x+y+z\\right)^2=0^2[\/tex] e, tamb\u00e9m, [tex]x^2+y^2+z^2=-2xy-2xz-2yz. \\, [\/tex] Com isso, segue que:<\/p>\n\n
\n[tex]x^4+y^4+z^4+2x^2y^2+2x^2z^2+2y^2z^2=4x^2y^2+4x^2z^2+4y^2z^2+8x^2yz+8xy^2z+8xyz^2 [\/tex]
\n[tex]x^4+y^4+z^4=2x^2y^2+2x^2z^2+2y^2z^2+8xyz\\underbrace{(x+y+z)}_{0}[\/tex]
\n[tex]x^4+y^4+z^4=2x^2y^2+2x^2z^2+2y^2z^2[\/tex]
\n[tex][x^4+y^4+z^4]+x^4+y^4+z^4=[x^4+y^4+z^4]+2x^2y^2+2x^2z^2+2y^2z^2[\/tex]
\n[tex]2x^4+2y^4+2z^4=x^4+y^4+z^4+2x^2y^2+2x^2z^2+2y^2z^2[\/tex]
\n[tex]\\boxed{2x^4+2y^4+2z^4=(x^2+y^2+z^2)^2} \\, .[\/tex]<\/li>\n<\/ul>\n
\n[tex]\\quad (a^6+b^6+c^6)^2=2(a^{12}+b^{12}+c^{12})=2\\times 8,[\/tex]
\ndonde, finalmente, [tex] \\, \\fcolorbox{black}{#eee0e5}{$a^6+b^6+c^6=4$} \\, .[\/tex]\n<\/p><\/div>\n<\/div>\n<\/div>\n
\n\n
\n[tex]\\boxed{\\sqrt[3]{3\\alpha^2 -4\\alpha \\, }}=\\sqrt[3]{3\\alpha^2 -3\\alpha – \\alpha \\, }=\\sqrt[3]{3\\alpha^2-3\\alpha+1-\\alpha^3 \\, }=\\sqrt[3]{(1-\\alpha)^3 \\, }=\\boxed{1-\\alpha} \\, .[\/tex]<\/li>\n
\nObserve, inicialmente, que se [tex]\\alpha^3-\\alpha-1=0 \\, [\/tex], ent\u00e3o [tex]\\alpha^3=\\alpha+1.[\/tex]
\n[tex]\\begin{align*}\\boxed{\\alpha \\sqrt[4]{2\\alpha^2 +3\\alpha +2}}&=\\alpha \\sqrt[4]{\\alpha^2+\\alpha^2+\\alpha+2\\alpha+2}=\\alpha \\sqrt[4]{\\alpha^2+\\alpha(\\alpha+1)+2(\\alpha+1)}\\\\
\n&=\\alpha \\sqrt[4]{\\alpha^2+\\alpha\\alpha^3+2\\alpha^3} =\\alpha \\sqrt[4]{\\alpha^2+\\alpha^4+2\\alpha^3}\\\\
\n&=\\alpha \\sqrt[4]{\\alpha^2+\\alpha^4+\\alpha^3+\\alpha^3}=\\alpha \\sqrt[4]{\\alpha^2(1+\\alpha)+\\alpha^3(1+\\alpha)}\\\\
\n&=\\alpha \\sqrt[4]{\\alpha^2\\alpha^3+\\alpha^3\\alpha^3}=\\alpha \\sqrt[4]{\\alpha^5+\\alpha^6}=\\alpha \\sqrt[4]{\\alpha^5(1+\\alpha)}\\\\
\n&=\\alpha \\sqrt[4]{\\alpha^5\\alpha^3}=\\alpha \\sqrt[4]{\\alpha^8}=\\alpha\\alpha^2=\\alpha^3=\\boxed{1+\\alpha} \\, .\\end{align*}[\/tex]<\/li>\n<\/ul>\n
\n[tex]\\qquad \\qquad \\sqrt[3]{3\\alpha^2 -4\\alpha}+\\alpha \\sqrt[4]{2\\alpha^2 +3\\alpha +2}=\\boxed{1-\\alpha}+\\boxed{1+\\alpha}= \\, \\fcolorbox{black}{#eee0e5}{$2$} \\, .[\/tex]\n<\/p><\/div>\n<\/div>\n<\/div>\n
\n\n
\n[tex]\\qquad \\dfrac{1}{1+a^k}+\\dfrac{1}{1+a^{-k}}=\\dfrac{1+a^{-k}+1+a^k}{(1+a^{-k})(1+a^k)}=\\dfrac{2+a^k+a^{-k}}{1+a^{-k}+a^k+1}=1,[\/tex]<\/p>\n
\n [tex]\\qquad \\qquad \\, \\fcolorbox{black}{#eee0e5}{$\\displaystyle\\sum_{k=-2007}^{k=2007}\\dfrac{1}{1+a^k}=2007+\\dfrac{1}{2}=\\dfrac{4015}{2}$} \\, .[\/tex]\n<\/p><\/div>\n<\/div>\n<\/div>\n
\n
\n[tex]\\qquad \\qquad \\dfrac{1}{\\sqrt{2+\\sqrt{2}}}+\\dfrac{1}{\\sqrt{6+\\sqrt{6}}} \\gt \\dfrac{5}{6}.[\/tex]<\/p>\n\n
\n[tex]\\qquad \\dfrac{1}{2}\\lt \\dfrac{1}{\\left(\\sqrt{2+\\sqrt{2}}\\right)};[\/tex][tex]\\qquad \\color{#800000}{(i)}[\/tex]
\n[tex]\\qquad \\dfrac{1}{3}\\lt \\dfrac{1} {\\left(\\sqrt{6+\\sqrt{6}}\\right)}.[\/tex][tex]\\qquad \\color{#800000}{(ii)}[\/tex]
\nDessa forma, adicionando [tex]\\color{#800000}{(i)}[\/tex] e [tex]\\color{#800000}{(ii)}[\/tex], vem que [tex]\\boxed{\\dfrac{5}{6}\\lt \\dfrac{1}{\\sqrt{2+\\sqrt{2}}}+\\dfrac{1}{\\sqrt{6+\\sqrt{6}}}}.[\/tex]\n<\/p><\/div>\n<\/div>\n<\/div>\n
\n
\n(a)<\/strong> [tex] \\sqrt{\\dfrac{1}{6}+\\dfrac{\\sqrt{5}}{18}}-\\sqrt{\\dfrac{1}{6}-\\dfrac{\\sqrt{5}}{18}}[\/tex];<\/p>\n
\n[tex]\\sqrt{\\dfrac{1}{6}+\\dfrac{\\sqrt{5}}{18}}-\\sqrt{\\dfrac{1}{6}-\\dfrac{\\sqrt{5}}{18}}=\\sqrt{\\dfrac{3+\\sqrt{5}}{18}}-\\sqrt{\\dfrac{3-\\sqrt{5}}{18}}=[\/tex]
\n[tex]=\\sqrt{\\dfrac{6+2\\sqrt{5}}{36}}-\\sqrt{\\dfrac{6-2\\sqrt{5}}{36}}=\\sqrt{\\left(\\dfrac{1+\\sqrt{5}}{6}\\right)^2}-\\sqrt{\\left(\\dfrac{\\sqrt{5}-1}{6}\\right)^2}= \\left|\\dfrac{1+\\sqrt{5}}{6}\\right|-\\left|\\dfrac{\\sqrt{5}-1}{6}\\right|=[\/tex]
\n[tex]=\\dfrac{1+\\sqrt{5}}{6}-\\dfrac{\\sqrt{5}-1}{6}=\\dfrac{2}{6}=\\dfrac{1}{3}.[\/tex]<\/p>\n
\n
\n
\n[tex] \\sqrt{1+\\dfrac{2}{5}}\\cdot\\sqrt{1+\\dfrac{2}{6}}\\cdot\\sqrt{1+\\dfrac{2}{7}}\\cdot\\sqrt{1+\\dfrac{2}{8}}\\cdots \\sqrt{1+\\dfrac{2}{57}}\\cdot\\sqrt{1+\\dfrac{2}{58}}=[\/tex]
\n[tex] \\sqrt{\\left(1+\\dfrac{2}{5}\\right)\\cdot \\left(1+\\dfrac{2}{6}\\right)\\cdot \\left(1+\\dfrac{2}{7}\\right)\\cdot \\left(1+\\dfrac{2}{8}\\right)\\cdot \\left(1+\\dfrac{2}{57}\\right)\\cdot \\left(1+\\dfrac{2}{58}\\right)}=[\/tex]
\n[tex]=\\sqrt{\\dfrac{7}{5}\\cdot \\dfrac{8}{6}\\cdot \\dfrac{9}{7}\\cdot \\dfrac{10}{8}\\cdot \\dfrac{11}{9}\\cdot \\ldots \\cdot \\dfrac{59}{57}\\cdot \\dfrac{60}{58}}=\\sqrt{\\dfrac{1}{5}\\cdot \\dfrac{1}{6}\\cdot 59\\cdot 60}=\\sqrt{118}.[\/tex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n
\n
\n[tex]\\quad 2^x.3^y=\\left(24^{\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{4}+\\cdots+\\frac{1}{60}}\\right).\\left(24^{\\frac{1}{3}+\\frac{1}{4}+\\frac{1}{5}+\\cdots+\\frac{1}{60}}\\right)^2.\\left(24^{\\frac{1}{4}+\\frac{1}{5}+\\frac{1}{6}+\\cdots+\\frac{1}{60}}\\right)^3 \\cdots \\left(24^{\\frac{1}{60}}\\right)^{59}.[\/tex]<\/p>\n
\n[tex]\\,
\n\\begin{align*}2^x.3^y &=\\left(24^{\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{4}+\\cdots+\\frac{1}{60}}\\right) \\cdot \\left(24^{\\frac{1}{3}+\\frac{1}{4}+\\frac{1}{5}+\\cdots+\\frac{1}{60}}\\right)^2 \\cdot \\left(24^{\\frac{1}{4}+\\frac{1}{5}+\\frac{1}{6}+\\cdots+\\frac{1}{60}}\\right)^3 \\, \\cdots \\, \\left(24^{\\frac{1}{60}}\\right)^{59}\\\\
\n&=24^{(\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{4}+\\cdots+\\frac{1}{60})} \\cdot 24^{(\\frac{2}{3}+\\frac{2}{4}+\\frac{2}{5}+\\cdots+\\frac{2}{60})} \\cdot 24^{(\\frac{3}{4}+\\frac{3}{5}+\\frac{3}{6}+\\cdots+\\frac{3}{60})} \\, \\cdots \\, 24^{\\frac{59}{60}}\\\\
\n&=24^{\\left[\\left(\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{4}+\\cdots+\\frac{1}{60}\\right)+
\n \\left(\\frac{2}{3}+\\frac{2}{4}+\\frac{2}{5}+\\cdots+\\frac{2}{60}\\right)+
\n \\left (\\frac{3}{4}+\\frac{3}{5}+\\frac{3}{6}+\\cdots+\\frac{3}{60}\\right)+ \\cdots +
\n \\frac{59}{60}\\right]}\\\\
\n&=24^k,\\end{align*}[\/tex]
\ncom
\n[tex]\\quad
\n\\begin{align*}k&=\\left(\\dfrac{1}{2}+\\dfrac{1}{3}+\\dfrac{1}{4}+\\cdots+\\dfrac{1}{60}\\right)+
\n \\left(\\dfrac{2}{3}+\\dfrac{2}{4}+\\dfrac{2}{5}+\\cdots+\\dfrac{2}{60}\\right)+
\n \\left (\\dfrac{3}{4}+\\dfrac{3}{5}+\\dfrac{3}{6}+\\cdots+\\dfrac{3}{60}\\right)+ \\cdots +
\n \\dfrac{59}{60}\\\\
\n&=\\dfrac{1}{2}+\\left(\\dfrac{1}{3}+\\dfrac{2}{3}\\right)+\\left(\\dfrac{1}{4}+\\dfrac{2}{4}+\\dfrac{3}{4}\\right)+\\left(\\dfrac{1}{5}+\\dfrac{2}{5}+\\dfrac{3}{5}+\\dfrac{4}{5}\\right)+\\cdots +\\left(\\dfrac{1}{60}+\\dfrac{2}{60}+\\cdots+\\dfrac{59}{60}\\right)\\\\
\n&=\\dfrac{1}{2}+1+\\dfrac{3}{2}+2+\\cdots+\\dfrac{59}{2}\\\\
\n&=\\dfrac{1}{2}+\\dfrac{2}{2}+\\dfrac{3}{2}+\\dfrac{4}{2}+\\cdots+\\dfrac{59}{2}\\\\
\n&=\\dfrac{1}{2}\\left(1+2+3+\\cdots + 59\\right)\\\\
\n&= \\dfrac{1}{2}\\cdot \\dfrac{(1+59)\\cdot 59}{2}\\\\
\n&=15\\times 59. \\end{align*} [\/tex]<\/p>\n
\nFinalmente, [tex]x+y=45\\cdot59+15\\cdot59=60\\cdot59=3540.[\/tex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n
\n
\n[tex]\\qquad \\qquad a^2b^2(a^2b^2+4)=2(a^6+b^6).[\/tex]
\nMostre que [tex]a[\/tex] e [tex]b[\/tex] n\u00e3o podem ser ambos racionais.<\/p>\n
\n[tex]\\quad a^2b^2(a^2b^2+4)=2(a^6+b^6)\\iff a^4b^4+4a^2b^2=2a^6+2b^6\\iff\\\\
\n\\quad \\iff a^4b^4=2a^6+2b^6-4a^2b^2\\iff\\\\
\n\\quad \\iff a^4b^4-2a^6=2b^6-4a^2b^2 \\iff a^4\\left(b^4-2a^2\\right)=2b^2\\left(b^4-2a^2\\right) \\iff \\\\
\n\\quad \\iff a^4\\left(b^4-2a^2\\right)-2b^2\\left(b^4-2a^2\\right)=0 \\iff \\left(a^4-2b^2\\right)\\left(b^4-2a^2\\right)=0.[\/tex]
\nLogo [tex]a^4-2b^2=0 \\, [\/tex] ou [tex]b^4-2a^2=0[\/tex]. Agora, veja ainda que:
\n[tex]\\quad a^4-2b^2=0\\iff a^4=2b^2 \\iff \\dfrac{a^2}{b}=\\pm\\sqrt{2};[\/tex]
\n[tex]\\quad b^4-2a^2=0\\iff b^4=2a^2 \\iff \\dfrac{b^2}{a}=\\pm\\sqrt{2}.[\/tex]
\nSendo assim, temos que [tex]\\dfrac{a^2}{b}=\\pm\\sqrt{2}; \\, [\/tex] ou [tex]\\dfrac{b^2}{a}=\\pm\\sqrt{2}, [\/tex] e, portanto, [tex]a[\/tex] e [tex]b[\/tex] n\u00e3o podem ser ambos racionais, pois, se o fossem, [tex]\\dfrac{a^2}{b}[\/tex] e [tex] \\dfrac{b^2}{a}[\/tex] tamb\u00e9m seriam e a\u00ed ter\u00edamos uma contradi\u00e7\u00e3o, j\u00e1 que [tex]\\sqrt{2} \\, [\/tex] e [tex] \\, -\\sqrt{2}[\/tex] s\u00e3o n\u00fameros irracionais.\n<\/div>\n<\/div>\n<\/div>\n
\n
\n[tex]\\qquad \\alpha=\\sqrt{13}+\\sqrt{10+2\\sqrt{13}} \\quad[\/tex] e [tex]\\quad\\beta=\\sqrt{5+2\\sqrt{3}}+\\sqrt{18-2\\sqrt{3}+2\\sqrt{65-26\\sqrt{3}}}[\/tex],
\nmostrar que [tex]\\alpha=\\beta[\/tex].<\/p>\n
\n[tex]\\qquad \\beta=\\sqrt{5+2\\sqrt{3} \\, }+\\sqrt{18-2\\sqrt{3}+2\\sqrt{65-26\\sqrt{3}} \\, }[\/tex]
\n[tex]\\qquad \\beta=\\sqrt{5+2\\sqrt{3} \\, }+\\sqrt{18-2\\sqrt{3}+2\\sqrt{13}\\sqrt{5-2\\sqrt{3}} \\, }[\/tex]
\n[tex]\\qquad \\beta=\\sqrt{5+2\\sqrt{3}}+\\sqrt{\\left(\\sqrt{13}+\\sqrt{5-2\\sqrt{3}}\\right)^2}[\/tex]
\n[tex]\\qquad \\beta=\\sqrt{5+2\\sqrt{3}}+\\sqrt{13}+\\sqrt{5-2\\sqrt{3} \\, }.[\/tex][tex]\\qquad \\color{#800000}{(i)}[\/tex]
\nAssim, segue por [tex]\\color{#800000}{(i)}[\/tex] que:
\n[tex]\\qquad \\begin{align*}\\left(\\beta-\\sqrt{13}\\right)^2 &=\\left(\\sqrt{5+2\\sqrt{3}}+\\sqrt{5-2\\sqrt{3}}\\right)^2\\\\
\n&=\\left(5+2\\sqrt{3}\\right)+2\\left(\\sqrt{5+2\\sqrt{3}}\\right)\\left(\\sqrt{5-2\\sqrt{3}}\\right)+\\left(5-2\\sqrt{3}\\right)\\\\
\n&=5+\\cancel{2\\sqrt{3}}+2\\left(\\sqrt{5+2\\sqrt{3}}\\right)\\left(\\sqrt{5-2\\sqrt{3}}\\right)+5-\\cancel{2\\sqrt{3}}\\\\
\n&=10+2\\left(\\sqrt{5+2\\sqrt{3}}\\right)\\left(\\sqrt{5-2\\sqrt{3}}\\right)\\\\
\n&=10+2\\sqrt{\\left(5+2\\sqrt{3}\\right)\\left(5-2\\sqrt{3}\\right)}\\\\
\n&=10+2\\sqrt{(5)^2-(2\\sqrt{3})^2}\\\\
\n&=10+2\\sqrt{25-12}\\\\
\n&=10+2\\sqrt{13},\\end{align*}[\/tex]
\ne assim, [tex]\\left(\\beta-\\sqrt{13}\\right)^2=10+2\\sqrt{13}.[\/tex][tex]\\qquad \\color{#800000}{(ii)}[\/tex]
\nPor outro lado, [tex]\\alpha-\\sqrt{13}=\\sqrt{10+2\\sqrt{13}}[\/tex], donde
\n[tex]\\qquad \\left(\\alpha-\\sqrt{13}\\right)^2=\\left(\\sqrt{10+2\\sqrt{13}}\\right)^2 =10+2\\sqrt{13}[\/tex]
\nou seja [tex]\\left(\\alpha-\\sqrt{13}\\right)^2=10+2\\sqrt{13}.[\/tex][tex]\\qquad \\color{#800000}{(iii)}[\/tex]
\nCom isso, por [tex]\\color{#800000}{(ii)} \\, [\/tex] e [tex] \\, \\color{#800000}{(iii)}[\/tex], [tex]\\left(\\alpha-\\sqrt{13}\\right)^2=\\left(\\beta-\\sqrt{13}\\right)^2 \\, [\/tex], e, ent\u00e3o:
\n[tex]\\qquad \\left(\\alpha-\\sqrt{13}\\right)^2=\\left(\\beta-\\sqrt{13}\\right)^2[\/tex]
\n[tex]\\qquad \\sqrt{\\left(\\alpha-\\sqrt{13}\\right)^2}=\\sqrt{\\left(\\beta-\\sqrt{13}\\right)^2}[\/tex]
\n[tex]\\qquad |\\alpha-\\sqrt{13}|=|\\beta-\\sqrt{13}|.[\/tex][tex]\\qquad \\color{#800000}{(iv)}[\/tex]
\nComo [tex]\\alpha-\\sqrt{13}=\\sqrt{10+2\\sqrt{13}} \\, \\gt 0 \\, [\/tex] e [tex] \\, \\beta-\\sqrt{13}=\\sqrt{5+2\\sqrt{3}}+\\sqrt{5-2\\sqrt{3}} \\, \\gt 0 \\, [\/tex], segue de [tex]\\color{#800000}{(iv)}[\/tex] que [tex]\\alpha-\\sqrt{13}=\\beta-\\sqrt{13}, \\, [\/tex] ou ainda, que [tex]\\boxed{\\alpha=\\beta} \\, .[\/tex]\n<\/div>\n<\/div>\n<\/div>\n
\n\n
\nMas, por outro lado, sabemos que determinadas manipula\u00e7\u00f5es alg\u00e9bricas podem ser feitas tanto para desigualdades como para igualdades, ent\u00e3o, indicando por [tex]\\boxed{?}[\/tex] qualquer um dos sinais [tex] \\, \\lt \\, , \\, \\gt \\, , \\, = \\, , \\, [\/tex] e observando que [tex]a \\gt 0, \\, [\/tex] [tex]a+1 \\gt 0 \\, [\/tex] e [tex]a-1 \\gt 0, \\, [\/tex] segue que:<\/p>\n
\nComo sabemos que [tex]a^2-1 \\lt a^2, \\, [\/tex] ent\u00e3o
\n[tex]\\qquad a^2-1 \\, \\lt \\, a^2 \\iff (a-1)(a+1) \\, \\lt \\, a^2 \\iff \\dfrac{a-1}{a} \\, \\lt \\, \\dfrac{a}{a+1}.[\/tex]
\nConsequentemente, das tr\u00eas op\u00e7\u00f5es, temos [tex] \\, \\dfrac{a-1}{a} \\, \\lt \\, \\dfrac{a}{a+1} \\, [\/tex] e, com isso, [tex] \\, \\dfrac{123457+10^{999}}{123458+10^{999}} \\, [\/tex] \u00e9 o maior dentre os dois n\u00fameros!\n<\/p><\/div>\n<\/div>\n<\/div>\n
\n
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