{"id":90570,"date":"2017-12-03T17:32:03","date_gmt":"2017-12-03T19:32:03","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=90570"},"modified":"2020-11-03T15:29:56","modified_gmt":"2020-11-03T17:29:56","slug":"malabarismos-aritmeticos-e-algebricos-problemas-3","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/malabarismos-aritmeticos-e-algebricos-problemas-3\/","title":{"rendered":"Malabarismos aritm\u00e9ticos e alg\u00e9bricos – Problemas (3)"},"content":{"rendered":"

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Identidades Alg\u00e9bricas <\/p>\n

Solu\u00e7\u00f5es dos problemas propostos – Terceira Sala<\/p>\n

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\nPara ajudar no entendimento das solu\u00e7\u00f5es, relacionamos algumas identidades apresentadas no nosso Blog. <\/p>\n

\n[tex]ax+ay=a(x+y)[\/tex]
\n[tex]a^2-b^2=(a+b)(a-b)[\/tex]
\n[tex]a^2\\pm 2ab+b^2=(a\\pm b)^2[\/tex]
\n[tex]a^3\\pm 3a^2b+3ab^2\\pm b^3=(a\\pm b)^3[\/tex]
\n[tex]a^2+b^2+c^2+2ab+2ac+2bc=(a+b+c)^2[\/tex]
\n[tex]a^3+b^3+c^3+3(a+b)(a+c)(b+c)=(a+b+c)^3[\/tex]
\n[tex]x^2+x(a+b)+ab=(x+a)(x+b)[\/tex]
\n[tex]x^3+(a+b+c)x^2+(ab+ac+bc)x+abc=(x+a)(x+b)(x+c)[\/tex]
\n[tex](ax+by)^2+(ay-bx)^2=(a^2+b^2)(x^2+y^2)[\/tex]
\n[tex]a^3\\pm b^3=(a\\pm b)(a^2\\mp ab+b^2)[\/tex]
\n[tex]a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)[\/tex]
\n[tex]a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2+\\dots+ab^{n-2}+b^{n-1})[\/tex], [tex]n[\/tex] natural
\n[tex]a^n+b^n=(a+b)(a^{n-1}-a^{n-2}b+a^{n-3}b^2-\\dots-ab^{n-2}+b^{n-1})[\/tex], [tex]n[\/tex] \u00edmpar
\n[tex]1+2+3+ \\cdots+t=\\dfrac{(1+t)\\cdot t}{2}[\/tex]\n<\/div>\n

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[tex]25.[\/tex] Sem efetuar as distributividades e os produtos, desenvolva a express\u00e3o [tex](a+b+c+d)^3[\/tex], admitindo que [tex]a, \\, b, \\, c, \\, d \\, [\/tex] s\u00e3o n\u00fameros reais.<\/p>\n

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\n <\/p>\n
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\nSabemos que <\/p>\n
    \n
  • [tex](a+b+c+d)^3=\\underbrace{(a+b+c+d)}_{I}\\cdot \\underbrace{(a+b+c+d)}_{II}\\cdot \\underbrace{(a+b+c+d)}_{III}[\/tex].<\/li>\n<\/ul>\n

    Inicialmente, observe que, se f\u00f4ssemos desenvolver a express\u00e3o dada, encontrar\u00edamos [tex]64[\/tex] parcelas, com uma das seguintes formas:
    \n [tex]\\qquad \\qquad \\boxed{x^3} \\quad [\/tex] , [tex]\\quad \\boxed{x^2y}\\quad [\/tex] e [tex]\\quad \\boxed{xyz} \\, \\, [\/tex], com [tex] \\, \\, x, \\, y, \\, z \\in \\{a, \\, b, \\, c, \\, d\\}[\/tex].
    \nAgora, observe que:<\/p>\n

      \n
    • Termos da forma [tex]\\boxed{x^3}[\/tex] aparecem quando multiplicamos [tex]x\\cdot x\\cdot x.[\/tex]
      \nAssim, vamos ter quatro parcelas desse tipo na nossa express\u00e3o: <\/p>\n
        \n
          \n
        • [tex]a^3 \\, \\, [\/tex], [tex] \\, \\, b^3 \\, \\, [\/tex], [tex] \\, \\, c^3 \\, \\, [\/tex] e [tex] \\, \\, d^3[\/tex].<\/li>\n<\/ul>\n<\/ul>\n<\/li>\n<\/ul>\n
            \n
          • Termos da forma [tex]\\boxed{x^2y}[\/tex] aparecem quando multiplicamos dois [tex]x[\/tex] e um [tex]y[\/tex], com [tex]x\\ne y[\/tex].
            \nNo nosso caso, temos quatro escolhas para [tex]y[\/tex], ent\u00e3o poderemos ter parcelas da forma [tex]x^2a, \\, x^2b, \\, x^2c, \\, x^2d[\/tex]. Mas, para cada um dos quatro termos [tex]x^2a, \\, x^2b, \\, x^2c, \\, x^2d \\, [\/tex], temos tr\u00eas escolhas para [tex]x[\/tex]; assim:<\/p>\n
              \n
                \n
              • para [tex]\\textcolor{red}{x}^2a[\/tex] obtemos os termos [tex]\\textcolor{red}{b}^2a[\/tex] , [tex]\\textcolor{red}{c}^2a[\/tex] e [tex]\\textcolor{red}{d}^2a[\/tex];<\/li>\n
              • para [tex]\\textcolor{blue}{x}^2b[\/tex] obtemos os termos: [tex]\\textcolor{blue}{a}^2b[\/tex] , [tex]\\textcolor{blue}{c}^2b[\/tex] , [tex]\\textcolor{blue}{d}^2b[\/tex];<\/li>\n
              • para [tex]\\textcolor{green}{x}^2c[\/tex] obtemos os termos: [tex]\\textcolor{green}{a}^2c[\/tex] , [tex]\\textcolor{green}{b}^2c[\/tex] , [tex]\\textcolor{green}{d}^2c[\/tex];<\/li>\n
              • para [tex]\\textcolor{#A020F0}{x}^2d[\/tex] obtemos os termos: [tex]\\textcolor{#A020F0}{a}^2d[\/tex] , [tex]\\textcolor{#A020F0}{b}^2d[\/tex] , [tex]\\textcolor{#A020F0}{c}^2d[\/tex].<\/li>\n<\/ul>\n<\/ul>\n<\/li>\n

                Mas perceba que, para construirmos os termos [tex]x^2y[\/tex], escolhido um [tex]\\textcolor{#8B0000}{y}[\/tex], esse [tex]\\textcolor{#8B0000}{y}[\/tex] est\u00e1 em cada um dos tr\u00eas par\u00eanteses [tex]I , II , III[\/tex], j\u00e1 que [tex]x\\cdot x\\cdot \\textcolor{#8B0000}{y}=x\\cdot \\textcolor{#8B0000}{y}\\cdot x=\\textcolor{#8B0000}{y} \\cdot x \\cdot x=x^2\\textcolor{#8B0000}{y} \\, [\/tex].
                \nAssim, na nossa express\u00e3o aparecer\u00e3o [tex]3\\times 12=36 \\, [\/tex] parcelas da forma [tex]x^2y[\/tex]:<\/p>\n

                  \n[tex] 3(\\textcolor{red}{b}^2a)+3(\\textcolor{red}{c}^2a)+3(\\textcolor{red}{d}^2a)+3(\\textcolor{blue}{a}^2b)+3(\\textcolor{blue}{c}^2b)+3(\\textcolor{blue}{d}^2b)+3(\\textcolor{green}{a}^2c)+3(\\textcolor{green}{b}^2c)+3(\\textcolor{green}{d}^2c)+3(\\textcolor{#A020F0}{a}^2d)+3(\\textcolor{#A020F0}{b}^2d)+3(\\textcolor{#A020F0}{c}^2d).[\/tex]<\/span><\/ul>\n

                  \n

                • Finalmente, termos da forma [tex]\\boxed{xyz}[\/tex] aparecem quando multiplicamos um [tex]x[\/tex], um [tex]y[\/tex] e um [tex]z[\/tex], com [tex]x \\ne y[\/tex], [tex]x \\ne z[\/tex] e [tex]y \\ne z[\/tex].
                  \nNa nossa express\u00e3o temos quatro grupos de tr\u00eas n\u00fameros distintos:
                  \n"[tex]a, \\, b, \\, c \\, [\/tex]" ; "[tex]a, \\, b, \\, d \\, [\/tex]" ; "[tex] \\, a, \\, c, \\, d \\, [\/tex]" ; "[tex]b, \\, c, \\, d \\, [\/tex]".
                  \nMas, como o produto de n\u00fameros reais \u00e9 comutativo, um mesmo termo da forma [tex]\\boxed{xyz}[\/tex] pode ser obtido de seis modos diferentes, j\u00e1 que temos tr\u00eas escolhas poss\u00edveis para [tex]x[\/tex] (uma em cada par\u00eantese), duas escolhas poss\u00edveis para [tex]y[\/tex] (uma em cada um dos dois par\u00eanteses restantes) e uma escolha para [tex]z[\/tex] (no par\u00eantese que sobrou sem elemento escolhido). Assim, cada grupo de letras produz seis parcelas na nossa express\u00e3o.<\/li>\n<\/ul>\n

                  Dessa forma:
                  \n[tex]\\quad \\begin{align*}(a+b+c+d)^3 =\\quad & a^3+b^3+c^3+d^3+\\\\
                  \n& 3(b^2a+c^2a+d^2a+a^2b+c^2b+d^2b+\\\\
                  \n& a^2c+b^2c+d^2c+a^2d+b^2d+c^2d)+\\\\
                  \n& 6(abc+abd+acd+bcd).\\end{align*}[\/tex]\n<\/p><\/div>\n<\/div>\n<\/div>\n

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                  [tex]26.[\/tex] (Mold\u00e1via<\/strong>) Os n\u00fameros inteiros [tex]x[\/tex], [tex]y[\/tex], [tex]z[\/tex] satisfazem a rela\u00e7\u00e3o [tex]x+y+z=0[\/tex].
                  \nMostre que o n\u00famero [tex]2(x^4+y^4+z^4)[\/tex] \u00e9 um quadrado perfeito.<\/p>\n

                  \n

                  \n <\/p>\n
                  \n
                  \n Dado que [tex]x+y+z=0[\/tex] , ent\u00e3o temos tamb\u00e9m que [tex](x+y+z)^2=0 \\, [\/tex], logo segue que:
                  \n[tex] \\qquad (x+y+z)\\cdot (x+y+z)=0\\\\
                  \n\\qquad x^2+xy+xz+yx+y^2+yz+zx+zy+z^2=0\\\\
                  \n\\qquad x^2+y^2+z^2+2xy+2xz+2yz=0[\/tex]
                  \ne, assim, temos que [tex] x^2+y^2+z^2=-2xy-2xz-2yz \\, [\/tex] , donde:
                  \n[tex]\\qquad \\left(x^2+y^2+z^2\\right)^2=\\left(-2xy-2xz-2yz\\right)^2[\/tex]
                  \n[tex]\\qquad \\left(x^2+y^2+z^2\\right)^2=\\left(2xy+2xz+2yz\\right)^2[\/tex]
                  \n[tex]\\qquad 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]\\qquad x^4+y^4+z^4=2x^2y^2+2x^2z^2+2y^2z^2+8xyz \\underbrace{(x+y+z)}_{0}[\/tex]
                  \n[tex]\\qquad x^4+y^4+z^4=2x^2y^2+2x^2z^2+2y^2z^2[\/tex]
                  \n[tex]\\qquad 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]\\qquad 2x^4+2y^4+2z^4=x^4+y^4+z^4+2x^2y^2+2x^2z^2+2y^2z^2[\/tex]
                  \n[tex]\\qquad 2x^4+2y^4+2z^4=(x^2+y^2+z^2)^2[\/tex]
                  \n[tex]\\qquad 2(x^4+y^4+z^4)=(x^2+y^2+z^2)^2.[\/tex]
                  \nA igualdade [tex]\\boxed{2(x^4+y^4+z^4)=(x^2+y^2+z^2)^2}[\/tex] mostra que, nas condi\u00e7\u00f5es do problema, o n\u00famero [tex]2(x^4+y^4+z^4)[\/tex] \u00e9 um quadrado perfeito.\n<\/div>\n<\/div>\n<\/div>\n

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                  [tex]27.[\/tex] (Rioplatense<\/strong>) Ache o valor da soma <\/p>\n

                  \n

                  [tex]\\,\\, \\sqrt{1+\\dfrac{1}{1^2}+\\dfrac{1}{2^2}}+\\sqrt{1+\\dfrac{1}{2^2}+\\dfrac{1}{3^2}}+\\sqrt{1+\\dfrac{1}{3^2}+\\dfrac{1}{4^2}}+…+\\sqrt{1+\\dfrac{1}{2005^2}+\\dfrac{1}{2006^2}}[\/tex].<\/p>\n

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                  \n <\/p>\n
                  \n
                  \nObserve que cada parcela da soma em quest\u00e3o \u00e9 da forma [tex]\\boxed{\\sqrt{1+\\dfrac{1}{a^2}+\\dfrac{1}{(a+1)^2}}},[\/tex] sendo [tex]a[\/tex] um n\u00famero natural n\u00e3o nulo. Assim, vamos trabalhar primeiramente nessa express\u00e3o.
                  \nPois bem, observe que, se [tex]a[\/tex] \u00e9 um n\u00famero natural n\u00e3o nulo, ent\u00e3o segue que:
                  \n[tex]\\begin{align*}\\qquad \\sqrt{1+\\dfrac{1}{a^2}+\\dfrac{1}{(a+1)^2}}&=\\sqrt{\\dfrac{a^2(a+1)^2+(a+1)^2+a^2}{a^2(a+1)^2}}\\\\
                  \n&=\\sqrt{\\dfrac{a^4+2a^3+a^2+a^2+2a+1+a^2}{a^2(a+1)^2}}\\\\
                  \n&=\\sqrt{\\dfrac{a^4+2a^3+3a^2+2a+1}{a^2(a+1)^2}}\\\\
                  \n&=\\sqrt{\\dfrac{(a^2+a+1)^2}{a^2(a+1)^2}}\\\\
                  \n&=\\dfrac{a^2+a+1}{a(a+1)} =\\dfrac{a(a+1)+1}{a(a+1)}\\\\
                  \n&=1+\\dfrac{1}{a(a+1)}=1+\\dfrac{1}{a}-\\dfrac{1}{a+1}.\\end{align*}[\/tex]
                  \nAssim, [tex] \\, \\, \\boxed{\\sqrt{1+\\dfrac{1}{a^2}+\\dfrac{1}{(a+1)^2}}}=\\boxed{1+\\dfrac{1}{a}-\\dfrac{1}{a+1}} \\, [\/tex], para todo n\u00famero natural n\u00e3o nulo [tex]a[\/tex].
                  \nPodemos, agora, aplicar essa igualdade para calcular o valor da express\u00e3o proposta no enunciado.<\/p>\n

                  \n

                  [tex] \\sqrt{1+\\dfrac{1}{1^2}+\\dfrac{1}{2^2}}+\\sqrt{1+\\dfrac{1}{2^2}+\\dfrac{1}{3^2}}+\\sqrt{1+\\dfrac{1}{3^2}+\\dfrac{1}{4^2}}+…+\\sqrt{1+\\dfrac{1}{2005^2}+\\dfrac{1}{2006^2}}=\\\\
                  \n\\qquad \\qquad =\\left(1+\\dfrac{1}{1}-\\dfrac{1}{2}\\right)+\\left(1+\\dfrac{1}{2}-\\dfrac{1}{3}\\right)+\\left(1+\\dfrac{1}{3}-\\dfrac{1}{4}\\right)+…+\\left(1+\\dfrac{1}{2005}-\\dfrac{1}{2006}\\right)\\\\
                  \n\\qquad \\qquad =2005+\\left(\\dfrac{1}{1}-\\dfrac{1}{2}\\right)+\\left(\\dfrac{1}{2}-\\dfrac{1}{3}\\right)+\\left(\\dfrac{1}{3}-\\dfrac{1}{4}\\right)+…+\\left(\\dfrac{1}{2005}-\\dfrac{1}{2006}\\right)\\\\
                  \n\\qquad \\qquad =2005+\\dfrac{1}{1}-\\dfrac{1}{2}+\\dfrac{1}{2}-\\dfrac{1}{3}+\\dfrac{1}{3}-\\dfrac{1}{4}+…+\\dfrac{1}{2005}-\\dfrac{1}{2006}\\\\
                  \n\\qquad \\qquad =2005+1-\\cancel{\\dfrac{1}{2}}+\\cancel{\\dfrac{1}{2}}-\\bcancel{\\dfrac{1}{3}}+\\bcancel{\\dfrac{1}{3}}-\\dots -\\cancel{\\dfrac{1}{2005}}+\\cancel{\\dfrac{1}{2005}}-\\dfrac{1}{2006}\\\\
                  \n\\qquad \\qquad =2006-\\dfrac{1}{2006}\\\\
                  \n\\qquad \\qquad =\\dfrac{2006^2-1}{2006}\\\\
                  \n\\qquad \\qquad =\\dfrac{2005\\cdot 2007}{2006} \\, .[\/tex]<\/p>\n

                  \n

                  Portanto,
                  \n [tex] \\, \\fcolorbox{black}{#eee0e5}{$\\sqrt{1+\\dfrac{1}{1^2}+\\dfrac{1}{2^2}}+\\sqrt{1+\\dfrac{1}{2^2}+\\dfrac{1}{3^2}}+\\sqrt{1+\\dfrac{1}{3^2}+\\dfrac{1}{4^2}}+…+\\sqrt{1+\\dfrac{1}{{2005}{^{2}}}+\\dfrac{1}{2006{^2}}}=\\dfrac{2005\\cdot 2007}{2006}$} \\, .[\/tex]\n<\/p><\/div>\n<\/div>\n<\/div>\n

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                  [tex]28.[\/tex] (Espanha-Adaptado<\/strong>) Sejam [tex]a[\/tex], [tex]b[\/tex], [tex]c[\/tex] n\u00fameros reais n\u00e3o nulos com [tex]a+b+c\\neq 0[\/tex] tais que<\/p>\n

                  \n

                  [tex]\\qquad \\qquad \\dfrac{1}{a}+\\dfrac{1}{b}+\\dfrac{1}{c}=\\dfrac{1}{a+b+c}[\/tex] .<\/p>\n

                  \n

                  Mostre que, para [tex]n[\/tex] \u00edmpar, se verifica a igualdade<\/p>\n

                  \n

                  [tex]\\qquad \\qquad \\dfrac{1}{a^n}+\\dfrac{1}{b^n}+\\dfrac{1}{c^n}=\\dfrac{1}{a^n+b^n+c^n}[\/tex] .<\/p>\n

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                  \n <\/p>\n
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