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3 years ago

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katrin2010 [14]3 years ago

7 0

$\dfrac{n^2+3n+2}{n^2+6n+8}-\dfrac{2n}{n+4}\\\\=\dfrac{n^2+2n+n+2}{n^2+4n+2n+8}-\dfrac{2n}{n+4}\\\\=\dfrac{n(n+2)+1(n+2)}{n(n+4)+2(n+4)}-\dfrac{2n}{n+4}\\\\=\dfrac{(n+2)(n+1)}{(n+2)(n+4)}-\dfrac{2n}{n+4}\\\\=\dfrac{n+1}{n+4}-\dfrac{2n}{n+4}=\dfrac{n+1-2n}{n+4}=\dfrac{1-n}{n+4}$

$Answer:\ \boxed{B.\ \dfrac{1-n}{n+4}}$

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