{"id":235425,"date":"2022-04-28T17:07:51","date_gmt":"2022-04-28T15:07:51","guid":{"rendered":"https:\/\/sherpas.com\/blog\/?p=235425"},"modified":"2025-09-29T13:36:56","modified_gmt":"2025-09-29T11:36:56","slug":"exercices-sur-les-fractions-rationnelles","status":"publish","type":"post","link":"https:\/\/sherpas.com\/blog\/exercices-sur-les-fractions-rationnelles\/","title":{"rendered":"Exercices sur les fractions rationnelles"},"content":{"rendered":"\n

Coinc\u00e9 sur un exercice portant sur les fractions rationnelles<\/strong> ?<\/p>\n

Ne laisse pas ce d\u00e9fi te freiner. Transforme tes difficult\u00e9s en succ\u00e8s avec des cours approfondis en math\u00e9matiques<\/a> sp\u00e9cifiquement ax\u00e9s sur les fractions rationnelles. \ud83c\udf93<\/p>\n

En te plongeant dans un apprentissage plus approfondi, tu pourras non seulement r\u00e9soudre l’exercice en question, mais aussi ma\u00eetriser pleinement ce concept complexe.<\/p>\n\n\n\n

Exercices d’application sur les fractions rationnelles<\/h2>\n\n\n\n

Exercice 1<\/span> : <\/strong><\/p>\n\n\n\n

\u23f0 Dur\u00e9e : 10 min<\/p>\n\n\n\n

\ud83d\udcaa Difficult\u00e9 : niveau 1\/3<\/p>\n\n\n\n\nTrouver la d\u00e9composition en \u00e9l\u00e9ments simples de la fraction rationnelle suivante :\n\n\n\n

\n
\n\n1. \"\dfrac{X^2+2X+5}{X^2-3X+2}\";\n<\/div>\n\n\n\n
\n\n2. \"\dfrac{X^3+1}{(X-1)^3}\".\n<\/div>\n<\/div>\n\n\n\n
<\/div>\n\n\n\n

Exercice 2<\/span> : <\/strong><\/p>\n\n\n\n

\u23f0 Dur\u00e9e : 10 min<\/p>\n\n\n\n

\ud83d\udcaa Difficult\u00e9 : niveau 2\/3<\/p>\n\n\n\n\nSoient \"n \in \mathbb{N}^*\" et \"P_n = 1 + X + \frac {X^2}{2!} + ... + \frac {X^n}{n!}\". Montrer que \"P_n\" n’admet que des racines complexes simples.\n\n\n\n

Corrig\u00e9s des exercices d’application sur les fractions rationnelles<\/h2>\n\n\n\n

Exercice 1<\/span> :<\/strong> <\/p>\n\n\n\n\n1. On note \"F = \frac{X^2+2X+5}{X^2 - 3X + 2}\".
\n

  • On d\u00e9termine les p\u00f4les de \"F\" qui sont les racines de \"X^2 - 3X + 2\" soit : 1 et 2. Ainsi : \n
    \n

      <\/span>   <\/span>\"\[<\/p><\/li>\n
    \n

  • On a \"deg(F) = 0\", on d\u00e9termine la partie enti\u00e8re de \"F : X^2+2X+5 = 1 \times (X^2 - 3X + 2) + 5X + 3\". On en d\u00e9duit qu’il existe \"(\alpha,\beta) \in \mathbb{C}^2\" tel que :\n
    \n

      <\/span>   <\/span>\"\[<\/p><\/li>\n
    \n

  • On d\u00e9termine les coefficients des p\u00f4les simples.\n
    – Pour le p\u00f4le 1, on multiplie la relation par \"X-1\" : \n
    \n\n

      <\/span>   <\/span>\"\[\frac{(X^2 + 2X + 5)(X-1)}{(X-1)(X-2)} = (X-1)(1+ \frac{\beta}{X-2}) + \alpha.\]\"<\/p>\n\n
    En \u00e9valuant en 1, on obtient \"\alpha = -8\".\n
    – Pour le p\u00f4le 2, on multiplie la relation par \"X-2\" :\n\n

      <\/span>   <\/span>\"\[\frac{(X^2 + 2X + 5)(X-2)}{(X-1)(X-2)} = (X-2)(1+ \frac{\alpha}{X-1}) + \beta.\]\"<\/p>\n\n
    En \u00e9valuant en 2, on obtient \"\beta = 13\".\n
    Finalement : \"F = 1 - \frac{8}{X-1} + \frac{13}{X-2}\". <\/li>\n2. On pose \"F=\dfrac{X^3+1}{(X-1)^3}\".
    \n

  • On d\u00e9termine les p\u00f4les de \"F\" qui sont les racines de \"(X-1)^3\". Il n’y a qu’un p\u00f4le multiple : \"1\". <\/li>\n
  • On a \"\deg F=0\", on d\u00e9termine la partie enti\u00e8re de \"F\" : \"X^3+1 =1\times(X-1)^3+3X^2+3X+2\". On en d\u00e9duit qu’il existe \"(\alpha,\beta,\gamma)\in\C^3\" tel que :\n

      <\/span>   <\/span>\"\[F=1+\dfrac{\alpha}{X-1}+\dfrac{\beta}{(X-1)^2}+\dfrac{\gamma}{(X-1)^3}.\]\"<\/p> <\/li>\n

  • On d\u00e9termine les coefficients des p\u00f4les multiples.
    \n– pour trouver \"\gamma\", on multiplie la relation par \"(X-1)^3\" :\n

      <\/span>   <\/span>\"\[X^3+1=(X-1)^3+\alpha(X-1)^2+\beta(X-1)+\gamma.\]\"<\/p>\nEn \u00e9valuant en \"1\", on obtient \"\gamma=2\".
    \n– pour trouver \"\beta\", on calcule \"F-\dfrac{2}{(X-1)^3}=\dfrac{X^2+X+1}{(X-1)^2}\", on a alors :\n

      <\/span>   <\/span>\"\[\dfrac{X^2+X+1}{(X-1)^2}=1+\dfrac{\alpha}{X-1}+\dfrac{\beta}{(X-1)^2}.\]\"<\/p>
    \n\n\nOn multiplie la relation par \"(X-1)^2\" :\n

      <\/span>   <\/span>\"\[X^2+X+1=(X-1)^2+\alpha(X-1)+\beta.\]\"<\/p>
    \n\nEn \u00e9valuant en \"1\", on obtient \"\beta=3\".
    \n– pour trouver \"\alpha\", on calcule \"F-\dfrac{2}{(X-1)^3}-\dfrac{3}{(X-1)^2}=\dfrac{X+2}{X-1}\", on a alors :\n

      <\/span>   <\/span>\"\[\dfrac{X+2}{X-1}=1+\dfrac{\alpha}{X-1}.\]\"<\/p>\n\n\nOn multiplie la relation par \"(X-1)\" :\n

      <\/span>   <\/span>\"\[X+2=(X-1)+\alpha.\]\"<\/p>\n\nEn \u00e9valuant en \"1\", on obtient \"\alpha=3\".
    \nFinalement : \"F=1+\dfrac{3}{X-1}+\dfrac{3}{(X-1)^2}+\dfrac{2}{(X-1)^3}\". <\/li>\n\n\n\n

    <\/div>\n\n\n\n

    Exercice 2<\/span> : <\/strong><\/p>\n\n\n\n\nSi \"n = 1\", le r\u00e9sultat est \u00e9vident.\nSoit \"n \in \mathbb{N}^*\", \"n\geq 2\". Soit \"\alpha\" une racine complexe de multiplicit\u00e9 au moins 2. Alors \"P_n(\alpha) = 0\" et \"P_n'(\alpha) = 0\". Or, il est clair que \"P_n'= P_{n-1}\", ainsi \"P_n = P_n' + \frac {X^n}{n!}\". On en d\u00e9duit donc que : \n

      <\/span>   <\/span>\"\[<\/p>\nFinalement, \"\alpha = 0\", or il est clair que \"0\" n’est pas une racine de \"P_n\".\n
    En conclusion, \"P_n\" n’admet pas de racine complexe de multiplicit\u00e9 au moins 2, donc \"P_n\" n’admet que des racines complexes simples.\n\n\n\n

    \n
    \n
    \"livre<\/figure>\n<\/div>\n\n\n\n
    <\/div>\n\n\n\n
    \n
    <\/div>\n
    \n

    Cet article est extrait de l’ouvrage Maths MPSI-MP2I. Tout-en-un : cours, m\u00e9thodes, entra\u00eenement et corrig\u00e9s <\/em>(\u00e9ditions Vuibert, juin 2021) \u00e9crit par E. Thomas, S. Bellec, G. Boutard. ISBN n\u00b09782311408720<\/em><\/p>\n\n <\/div>\n <\/section>\n<\/div>\n<\/div>\n\n\n

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    \n 5\/5 - (2 votes) <\/div>\n <\/div>\n","protected":false},"excerpt":{"rendered":"

    Coinc\u00e9 sur un exercice portant sur les fractions rationnelles ? Ne laisse pas ce d\u00e9fi te freiner. Transforme (…)<\/p>\n","protected":false},"author":158,"featured_media":235610,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":true,"footnotes":""},"category":[803,810],"tag":[78,345],"class_list":["post-235425","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-apprendre-matiere","category-maths","tag-prepa","tag-prepa-scientifique"],"acf":[],"_links":{"self":[{"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/posts\/235425","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/users\/158"}],"replies":[{"embeddable":true,"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/comments?post=235425"}],"version-history":[{"count":0,"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/posts\/235425\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/media\/235610"}],"wp:attachment":[{"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/media?parent=235425"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/category?post=235425"},{"taxonomy":"tag","embeddable":true,"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/tag?post=235425"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}