Răspuns:
[tex]a) a=\frac{n}{n+1} +\frac{1}{n+1} \\\frac{n+1}{n+1} =1\\\\c)c=\frac{4n+9}{2n+6} -\frac{n}{n+3} -\frac{3}{2n+6} \\\frac{4n+9}{2(n+3)} -\frac{n}{n+3}-\frac{3}{2(n+3)} \\\frac{4n+9-2n-3}{2(n+3)} \\\frac{2n+6}{2(n+3)} \\ \frac{2(n+3)}{2(n+3)} =1\\\\e) e=\frac{-n+7}{3n+4} -\frac{-4n+3}{3n+4} \\\frac{-n+7-(-4n+3)}{3n+4} \\\frac{-n+7+4n-3}{3n+4} \\\frac{3n+4}{3n+4} =1[/tex]
Coroana?
Explicație pas cu pas:
