Sempre que for poss\u00edvel, simplifique as frac\u00e7\u00f5es e indique o dom\u00ednio da fun\u00e7\u00e3o.<\/p>\n
Aprecie a correc\u00e7\u00e3o dos resultados recorrendo \u00e0 calculadora gr\u00e1fica.<\/p>\n
- \n
- $f(x)=\\frac{2{{x}^{3}}-8{{x}^{2}}+8x}{{{x}^{3}}-4x}$;<\/li>\n
- $f(x)=\\frac{3{{x}^{2}}+5x-8}{-{{x}^{2}}-x+2}$;<\/li>\n
- $f(x)=\\frac{4{{x}^{3}}-3{{x}^{2}}+4x-3}{4x-3}$;<\/li>\n
- $f(x)=\\frac{{{x}^{2}}+x-6}{{{x}^{3}}-2{{x}^{2}}-x+2}$;<\/li>\n
- $f(x)=\\frac{{{x}^{2}}+2x-8}{{{x}^{3}}-8}$.<\/li>\n<\/ol>\nResolu\u00e7\u00e3o >><\/a><\/span><\/div><\/div>\n\n\nResolu\u00e7\u00e3o<\/b><\/span><\/p>\n
- \n
- ${{D}_{f}}=\\left\\{ x\\in \\mathbb{R}:{{x}^{3}}-4x\\ne 0 \\right\\}=\\left\\{ x\\in \\mathbb{R}:x({{x}^{2}}-4)\\ne 0 \\right\\}=\\mathbb{R}\\backslash \\left\\{ -2,0,2 \\right\\}$.\n
\\[\\frac{2{{x}^{3}}-8{{x}^{2}}+8x}{{{x}^{3}}-4x}=\\frac{2x({{x}^{2}}-4x+4)}{x({{x}^{2}}-4)}=\\frac{2x{{(x-2)}^{2}}}{x(x+2)(x-2)}=\\frac{2(x-2)}{x+2}=\\frac{2x-4}{x+2}\\]
\nSimplifica\u00e7\u00e3o v\u00e1lida\u00a0em $\\mathbb{R}\\backslash \\left\\{ -2,0,2 \\right\\}$.<\/p>\n![\"\"](\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/02\/11-pag-187-19-1.jpg\" "\\"J1\\"")
<\/a>\u00a0 ![\"\"](\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/02\/11-pag-187-19-2.jpg\" "\\"G1\\"")
<\/a>\u00a0 ![\"\"](\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/02\/11-pag-187-19-3.jpg\" "\\"G2\\"")
<\/a><\/p>\n![\"\"](\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/02\/11-pag-187-19-4.jpg\" "\\"G3\\"")
<\/a>\u00a0 ![\"\"](\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/02\/11-pag-187-19-5.jpg\" "\\"G4\\"")
<\/a><\/p>\n<\/li>\n- ${{D}_{f}}=\\left\\{ x\\in \\mathbb{R}:-{{x}^{2}}-x+2\\ne 0 \\right\\}=\\mathbb{R}\\backslash \\left\\{ -2,1 \\right\\}$.\n
\\[\\frac{3{{x}^{2}}+5x-8}{-{{x}^{2}}-x+2}=\\frac{(x-1)(3x+8)}{-(x+2)(x-1)}=\\frac{3x+8}{-x-2}\\]
\nSimplifica\u00e7\u00e3o v\u00e1lida em $\\mathbb{R}\\backslash \\left\\{ -2,1 \\right\\}$.<\/p>\nC\u00e1lculos auxiliares<\/strong>:
\n\\[-{{x}^{2}}-x+2=0\\Leftrightarrow x=\\frac{1\\mp \\sqrt{1+8}}{-2}\\Leftrightarrow x=-2\\vee x=1\\]
\n$\\begin{matrix}
\n{} & 3 & 5 & -8\u00a0 \\\\
\n1 & {} & 3 & 8\u00a0 \\\\
\n{} & 3 & 8 & 0\u00a0 \\\\
\n\\end{matrix}$
\n(Regra de Ruffini)<\/p>\n![\"\"](\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/02\/11-pag-187-19-6.jpg\" "\\"J2\\"")
<\/a>\u00a0 ![\"\"](\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/02\/11-pag-187-19-7.jpg\" "\\"G5\\"")
<\/a>\u00a0 ![\"\"](\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/02\/11-pag-187-19-8.jpg\" "\\"G6\\"")
<\/a><\/p>\n<\/li>\n- \n
${{D}_{f}}=\\left\\{ x\\in \\mathbb{R}:4x-3\\ne 0 \\right\\}=\\mathbb{R}\\backslash \\left\\{ \\frac{3}{4} \\right\\}$.
\n\\[\\frac{4{{x}^{3}}-3{{x}^{2}}+4x-3}{4x-3}=\\frac{(4x-3)({{x}^{2}}+1)}{4x-3}={{x}^{2}}+1\\]
\nSimplifica\u00e7\u00e3o v\u00e1lida em $\\mathbb{R}\\backslash \\left\\{ \\frac{3}{4} \\right\\}$.<\/p>\n![\"\"](\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/02\/11-pag-187-19-9.jpg\" "\\"J3\\"")
<\/a>\u00a0 ![\"\"](\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/02\/11-pag-187-19-10.jpg\" "\\"G7\\"")
<\/a>\u00a0 ![\"\"](\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/02\/11-pag-187-19-11.jpg\" "\\"G8\\"")
<\/a><\/p>\n<\/li>\n- \n
${{D}_{f}}=\\left\\{ x\\in \\mathbb{R}:{{x}^{3}}-2{{x}^{2}}-x+2 \\right\\}=\\left\\{ x\\in \\mathbb{R}:(x-1)(x-2)(x+1) \\right\\}=\\mathbb{R}\\backslash \\left\\{ -1,1,2 \\right\\}$.
\n\\[\\frac{{{x}^{2}}+x-6}{{{x}^{3}}-2{{x}^{2}}-x+2}=\\frac{(x+3)(x-2)}{(x-1)(x-2)(x+1)}=\\frac{x+3}{{{x}^{2}}-1}\\]
\nSimplifica\u00e7\u00e3o v\u00e1lida em $\\mathbb{R}\\backslash \\left\\{ -1,1,2 \\right\\}$.<\/p>\nC\u00e1lculos auxiliares<\/strong>:
\n\\[{{x}^{2}}+x-6=0\\Leftrightarrow x=\\frac{-1\\mp \\sqrt{1+24}}{2}\\Leftrightarrow x=-3\\vee x=2\\]
\n$\\begin{matrix}
\n{} & 1 & -2 & -1 & 2\u00a0 \\\\
\n1 & {} & 1 & -1 & -2\u00a0 \\\\
\n{} & 1 & -1 & -2 & 0\u00a0 \\\\
\n2 & {} & 2 & 2 & {}\u00a0 \\\\
\n{} & 1 & 1 & 0 & {}\u00a0 \\\\
\n\\end{matrix}$
\n(Regra de Ruffini)<\/p>\n![\"\"](\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/02\/11-pag-187-19-12.jpg\" "\\"J4\\"")
<\/a>\u00a0 ![\"\"](\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/02\/11-pag-187-19-13.jpg\" "\\"G9\\"")
<\/a>\u00a0 ![\"\"](\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/02\/11-pag-187-19-14.jpg\" "\\"G10\\"")
<\/a><\/p>\n<\/li>\n- \n
${{D}_{f}}=\\left\\{ x\\in \\mathbb{R}:{{x}^{3}}-8\\ne 0 \\right\\}=\\mathbb{R}\\backslash \\left\\{ 2 \\right\\}$.
\n\\[\\frac{{{x}^{2}}+2x-8}{{{x}^{3}}-8}=\\frac{(x+4)(x-2)}{(x-2)({{x}^{2}}+2x+4)}=\\frac{x+4}{{{x}^{2}}+2x+4}\\]
\nSimplifica\u00e7\u00e3o v\u00e1lida em $\\mathbb{R}\\backslash \\left\\{ 2 \\right\\}$.<\/p>\nC\u00e1lculos auxiliares<\/strong>:
\n\\[{{x}^{2}}+2x-8=0\\Leftrightarrow x=\\frac{-2\\mp \\sqrt{4+32}}{2}\\Leftrightarrow x=-4\\vee x=2\\]
\n$\\begin{matrix}
\n{} & 1 & 0 & 0 & -8\u00a0 \\\\
\n2 & {} & 2 & 4 & 8\u00a0 \\\\
\n{} & 1 & 2 & 4 & 0\u00a0 \\\\
\n\\end{matrix}$
\n(Regra de Ruffini)<\/p>\n - ${{D}_{f}}=\\left\\{ x\\in \\mathbb{R}:-{{x}^{2}}-x+2\\ne 0 \\right\\}=\\mathbb{R}\\backslash \\left\\{ -2,1 \\right\\}$.\n
- ${{D}_{f}}=\\left\\{ x\\in \\mathbb{R}:{{x}^{3}}-4x\\ne 0 \\right\\}=\\left\\{ x\\in \\mathbb{R}:x({{x}^{2}}-4)\\ne 0 \\right\\}=\\mathbb{R}\\backslash \\left\\{ -2,0,2 \\right\\}$.\n
![\"\"](\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/02\/11-pag-187-19-15.jpg\" "\\"J5\\"")
<\/a>\u00a0 ![\"\"](\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/02\/11-pag-187-19-16.jpg\" "\\"G11\\"")
<\/a>\u00a0 ![\"\"](\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/02\/11-pag-187-19-17.jpg\" "\\"G12\\"")
<\/a><\/p>\n<\/li>\n<\/ol>\n