AQUI<\/strong><\/a>; por ora, vamos relembrar essas rela\u00e7\u00f5es:<\/p>\n\nSe [tex]y_1[\/tex] e [tex]y_2[\/tex] s\u00e3o as ra\u00edzes da equa\u00e7\u00e3o do segundo grau [tex]\\, ay^2+by+c=0[\/tex], ent\u00e3o:
\n[tex]\\qquad \\qquad \\qquad \\boxed{y_1+y_2=\\dfrac{-b}{2a}}\\quad [\/tex] e [tex]\\quad \\boxed{y_1\\times y_2=\\dfrac{c}{2a}}\\,.[\/tex]<\/p>\n
Vamos utilizar essas duas rela\u00e7\u00f5es na discuss\u00e3o da equa\u00e7\u00e3o 1<\/strong><\/em>.<\/p>\n\n- Note que, se [tex]t\\ne1[\/tex], a equa\u00e7\u00e3o 1<\/strong><\/em> \u00e9 uma equa\u00e7\u00e3o do segundo grau, logo a soma das ra\u00edzes dessa equa\u00e7\u00e3o \u00e9
\n [tex]\\qquad\\dfrac{-(t^3-t)}{1-t}=\\dfrac{t-t^3}{1-t}=\\dfrac{t(1-t^2)}{1-t}=\\dfrac{t(1-t)(1+t)}{1-t}=\\boxed{t+t^2}[\/tex]
\ne o produto dessas ra\u00edzes \u00e9
\n [tex]\\qquad\\dfrac{\\, t^3-t^4}{1-t}=\\dfrac{t^3(1-t)}{1-t}=\\boxed{t^3}.[\/tex]
\nAssim, as ra\u00edzes da equa\u00e7\u00e3o 1<\/strong><\/em> s\u00e3o [tex]t[\/tex] e [tex]t^2[\/tex].
\nAnalisemos as duas possibilidades, lembrando que [tex]\\,\\boxed{ x^2=y}\\, [\/tex] e [tex]\\,\\boxed{\\sqrt{x-1}=t}\\, [\/tex].
\n i)<\/strong> Se [tex]y=t^2[\/tex], ent\u00e3o [tex]x^2=t^2~[\/tex] e, portanto, segue que:
\n[tex]\\qquad \\qquad \\sqrt{x-1}=t \\\\
\n\\qquad \\qquad \\left(\\sqrt{x-1}\\right)^2=t^2 \\\\
\n\\qquad \\qquad x-1=x^2 \\\\
\n\\qquad \\qquad x^2-x+1=0.[\/tex]
\n Mas a equa\u00e7\u00e3o [tex] x^2-x+1=0 [\/tex] n\u00e3o tem ra\u00edzes reais; logo [tex]y=t^2[\/tex] n\u00e3o define ra\u00edzes para o nosso problema.
\n ii)<\/strong> Se [tex]y=t[\/tex], ent\u00e3o [tex]x^2=t~[\/tex] e, agora, segue que:
\n[tex]\\qquad \\qquad \\sqrt{x-1}=t \\\\
\n\\qquad \\qquad \\left(\\sqrt{x-1}\\right)^2=t^2 \\\\
\n\\qquad \\qquad x-1=x^4 \\\\
\n\\qquad \\qquad x^4-x+1=0.[\/tex]
\n Analisemos, ent\u00e3o, a equa\u00e7\u00e3o [tex] x^4+1=x[\/tex].
\n ✔ Como [tex] x^4\\ge0[\/tex] [tex], x^4+1\\ge1[\/tex], assim [tex] x\\ge1[\/tex].
\n Dessa forma, [tex]x^4\\ge x^3 \\ge x^2\\ge x~[\/tex] e [tex]~x^4+1>x^4\\ge x[\/tex], donde [tex]x^4+1> x[\/tex].
\n Pelo exposto, [tex] x^4+1\\ne x[\/tex] e, portanto, [tex]y=t[\/tex] tamb\u00e9m n\u00e3o define ra\u00edzes para o nosso problema.<\/li>\n- Resta, apenas, a possibilidade de [tex]t=1[\/tex].
\nObserve que se [tex]t=1[\/tex], como [tex]\\sqrt{x-1}=t[\/tex], ent\u00e3o [tex](\\sqrt{x-1})^2=1^2[\/tex], donde [tex]x-1=1[\/tex].<\/li>\n<\/ul>\nAssim, [tex]x=2[\/tex] \u00e9 a \u00fanica solu\u00e7\u00e3o da equa\u00e7\u00e3o proposta<\/strong>.<\/p>\n
\nSolu\u00e7\u00e3o elaborada pelos Moderadores do Blog<\/strong> .<\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"Problema (Indicado a partir do 9\u00ba ano do E. F.) (ARGENTINA) Determinar todas as solu\u00e7\u00f5es reais da equa\u00e7\u00e3o [tex]\\qquad\\qquad \\dfrac{x^2}{x-1}+\\sqrt{x-1}+\\dfrac{\\sqrt{x-1}}{x^2}=\\dfrac{x-1}{x^2}+\\dfrac{1}{\\sqrt{x-1}}+\\dfrac{x^2}{\\sqrt{x-1}}.[\/tex] Solu\u00e7\u00e3o Fazendo [tex]\\,\\boxed{ x^2=y}\\, [\/tex] e [tex]\\,\\boxed{\\sqrt{x-1}=t}\\, [\/tex], podemos reescrever a equa\u00e7\u00e3o dada como [tex]\\qquad \\dfrac{y}{t^2}+t+\\dfrac{t}{y}=\\dfrac{t^2}{y}+\\dfrac{1}{t}+\\dfrac{y}{t} [\/tex] ou ainda [tex]\\qquad y^2(1-t)+y(t^3-t)+t^3-t^4=0 [\/tex]. Observe que obtivemos uma equa\u00e7\u00e3o do segundo grau em [tex]y[\/tex]: [tex]\\qquad (1-t)y^2+(t^3-t)y+(t^3-t^4)=0.\\quad [\/tex] … <\/p>\n
Continue lendo<\/a><\/p>\n","protected":false},"author":164,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"template-onecolumn.php","meta":{"footnotes":""},"class_list":["post-10487","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/10487","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/users\/164"}],"replies":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/comments?post=10487"}],"version-history":[{"count":69,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/10487\/revisions"}],"predecessor-version":[{"id":182691,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/10487\/revisions\/182691"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=10487"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
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