[tex]a)\:F(x)=\frac{x^3+x^2-9x-9}{x^3-9x}=\frac{x^2(x+1)-9(x+1)}{x(x^2-9)}=\frac{(x+1)(x^2-9)}{x(x^2-9)}=\frac{x+1}{x}\\
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[/tex]
La b) putem folosi rezultatul de la a) pentru un calcul rapid:
[tex]F(a)=a+1\Leftrightarrow\frac{a+1}{a}=a+1\Leftrightarrow a+1=a(a+1)\Leftrightarrow a+1=a^2+a\Leftrightarrow\\
a^2=1\Leftrightarrow |a|=1\Leftrightarrow a_{1,2}=\pm1\in \mathbb{R}\\
[/tex]
La c) folosim tot a):
[tex]S=F(6)+F(12)+F(20)+F(30)+F(42)+F(56)=\\
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=
\frac{6+1}{6}+\frac{12+1}{12}+\frac{20+1}{20}+\frac{30+1}{30}+\frac{42+1}{42}+\frac{56+1}{56}=\\
\\
=1+\frac{1}{6}+1+\frac{1}{12}+1+\frac{1}{20}+1+\frac{1}{42}+1+\frac{1}{56}=\\
\\aducem\:la\:acelasi\:numitor, 840\\
\\
=6+\frac{140+70+42+28+20+15}{840}=\\
\\
=6+\frac{315}{840}\\
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simplificam\:fractia\:prin\:105\\
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=6+\frac{3}{8}=\frac{51}{8}[/tex]
