Giải đáp: $\begin{array}{l}\dfrac{1}{{x + 2}} + \dfrac{3}{{{x^2} – 4}} + \dfrac{{x – 14}}{{{{\left( {x + 2} \right)}^2}\left( {x – 2} \right)}}\\ = \dfrac{1}{{x + 2}} + \dfrac{3}{{\left( {x – 2} \right)\left( {x + 2} \right)}} + \dfrac{{x – 14}}{{{{\left( {x + 2} \right)}^2}\left( {x – 2} \right)}}\\ = \dfrac{{\left( {x – 2} \right)\left( {x + 2} \right) + 3.\left( {x + 2} \right) + x – 14}}{{{{\left( {x + 2} \right)}^2}\left( {x – 2} \right)}}\\ = \dfrac{{{x^2} – 4 + 3x + 6 + x – 14}}{{{{\left( {x + 2} \right)}^2}\left( {x – 2} \right)}}\\ = \dfrac{{{x^2} + 4x – 12}}{{{{\left( {x + 2} \right)}^2}\left( {x – 2} \right)}}\\ = \dfrac{{\left( {x – 2} \right)\left( {x + 6} \right)}}{{{{\left( {x + 2} \right)}^2}\left( {x – 2} \right)}}\\ = \dfrac{{x + 6}}{{{{\left( {x + 2} \right)}^2}}}\end{array}$ Trả lời
\dfrac{1}{{x + 2}} + \dfrac{3}{{{x^2} – 4}} + \dfrac{{x – 14}}{{{{\left( {x + 2} \right)}^2}\left( {x – 2} \right)}}\\
= \dfrac{1}{{x + 2}} + \dfrac{3}{{\left( {x – 2} \right)\left( {x + 2} \right)}} + \dfrac{{x – 14}}{{{{\left( {x + 2} \right)}^2}\left( {x – 2} \right)}}\\
= \dfrac{{\left( {x – 2} \right)\left( {x + 2} \right) + 3.\left( {x + 2} \right) + x – 14}}{{{{\left( {x + 2} \right)}^2}\left( {x – 2} \right)}}\\
= \dfrac{{{x^2} – 4 + 3x + 6 + x – 14}}{{{{\left( {x + 2} \right)}^2}\left( {x – 2} \right)}}\\
= \dfrac{{{x^2} + 4x – 12}}{{{{\left( {x + 2} \right)}^2}\left( {x – 2} \right)}}\\
= \dfrac{{\left( {x – 2} \right)\left( {x + 6} \right)}}{{{{\left( {x + 2} \right)}^2}\left( {x – 2} \right)}}\\
= \dfrac{{x + 6}}{{{{\left( {x + 2} \right)}^2}}}
\end{array}$