Respuesta :

[tex]\bf \cfrac{n^2+3n+21}{n^2+6n+8}-\cfrac{2n}{n+4}\implies \cfrac{n^2+3n+21}{(n+4)(n+2)}-\cfrac{2n}{n+4} \\\\\\ \textit{so our LCD will just be (n+4)(n+2)} \\\\\\ \cfrac{n^2+3n+21~~~~-~~~~(n+2)(2n)}{(n+4)(n+2)} \\\\\\ \cfrac{n^2+3n+21~~~~-~~~~(2n^2+4n)}{(n+4)(n+2)} \\\\\\ \cfrac{n^2+3n+21~~~~-~~~~2n^2-4n}{(n+4)(n+2)} \implies \cfrac{-n^2-n+21}{(n+4)(n+2)}[/tex]

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