Toán Lớp 8: (1/x+2)+(3/x^2-4)+(x-14/(x+2)^2.(x-2))
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Toán Lớp 8: (1/x+2)+(3/x^2-4)+(x-14/(x+2)^2.(x-2))
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\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}$