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Problema Considere que [tex]a[\/tex] e [tex]b[\/tex] s\u00e3o dois n\u00fameros reais tais que [tex]0\\lt a\\lt b[\/tex] e [tex]a^2+b^2=4ab[\/tex]. \n<\/div>\n \n Solu\u00e7\u00e3o <\/p>\n [tex]\\qquad \\qquad a^2+b^2- 2ab=4ab -2ab \\\\ [tex]\\qquad \\qquad a^2+b^2+ 2ab=4ab +2ab \\\\ Substituindo [tex]\\textcolor{#800000}{(ii)}[\/tex] e [tex]\\textcolor{#800000}{(iv)}[\/tex] na express\u00e3o [tex] \\dfrac {a-b}{a+b}[\/tex], obtemos: \n<\/td>\n<\/tr>\n<\/table>\n 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.) Considere que [tex]a[\/tex] e [tex]b[\/tex] s\u00e3o dois n\u00fameros reais tais que [tex]0\\lt a\\lt b[\/tex] e [tex]a^2+b^2=4ab[\/tex]. Determine o valor da express\u00e3o [tex] \\dfrac {a-b}{a+b}.[\/tex] Solu\u00e7\u00e3o Subtraindo [tex] 2ab [\/tex] de ambos os membros da igualdade [tex]a^2+b^2=4ab[\/tex], segue que: [tex]\\qquad \\qquad a^2+b^2- 2ab=4ab -2ab \\\\ \\qquad … <\/p>\n
\n (Indicado a partir do 9\u00ba ano do E. F.)<\/span><\/p>\n
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\nDetermine o valor da express\u00e3o [tex] \\dfrac {a-b}{a+b}.[\/tex]<\/p>\n
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\n\\qquad \\qquad a^2 – 2ab + b^2 = 2ab \\\\
\n\\qquad \\qquad (a – b)^2 = 2ab\\\\
\n\\qquad \\qquad \\sqrt{(a – b)^2} = \\sqrt{2ab}\\\\
\n\\qquad \\qquad |a – b| = \\sqrt{2ab}.\\qquad \\textcolor{#800000}{(i)}[\/tex]
\nComo [tex] a [\/tex] e [tex] b [\/tex] s\u00e3o reais positivos com [tex] b\\gt a[\/tex] , temos que [tex] a – b \\lt 0[\/tex]. Logo, segue de [tex]\\textcolor{#800000}{(i)}[\/tex] que:
\n[tex]\\qquad \\qquad |a – b| = \\sqrt{2ab}\\\\
\n\\qquad \\qquad -\\left(a – b\\right)=\\sqrt{2ab}\\\\
\n\\qquad \\qquad \\boxed{a-b =-\\sqrt {2ab}} \\, .\\qquad \\textcolor{#800000}{(ii)} [\/tex]<\/p>\n
\n\\qquad \\qquad a^2+ 2ab + b^2 =6ab \\\\
\n\\qquad \\qquad (a+b)^2 =6ab\\\\
\n\\qquad \\qquad \\sqrt{(a+b)^2} = \\sqrt{6ab}\\\\
\n\\qquad \\qquad |a+b| = \\sqrt{6ab}.\\qquad \\textcolor{#800000}{(iii)}[\/tex]
\nComo [tex] a [\/tex] e [tex] b [\/tex] s\u00e3o reais positivos, conclu\u00edmos de [tex]\\textcolor{#800000}{(iii)}[\/tex] que:
\n[tex]\\qquad \\qquad |a+b| = \\sqrt{6ab}\\\\
\n\\qquad \\qquad \\boxed{a+b=\\sqrt{6ab}} \\, .\\qquad \\textcolor{#800000}{(iv)} [\/tex]<\/ul>\n
\n[tex]\\qquad \\qquad \\dfrac {a-b}{a+b} = \\dfrac {- \\sqrt {2ab}}{\\sqrt {6ab} }\\\\
\n\\qquad \\qquad \\dfrac {a-b}{a+b} = – \\sqrt {\\dfrac {2ab}{6ab}}\\\\
\n\\qquad \\qquad \\dfrac {a-b}{a+b} = – \\dfrac {1} {\\sqrt{3}} = -\\dfrac {\\sqrt {3}} {3} \\, .[\/tex]
\nPortanto, o valor da express\u00e3o \u00e9 [tex] \\, \\fcolorbox{black}{#eee0e5}{$- \\dfrac {\\sqrt{3}}{3}$} \\, .[\/tex]<\/p>\n
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