a)1/x+2 + 2/2-x + x/x^2-4 b) 3x^2+5x+1/x^3-1 – 1-x/x^2+x+1 – 3/x-1 quy đồg

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1. Đáp án:

   $\begin{array}{l}
   a)\,\,{\frac{{ – 2}}{{x – 2}}}\\
   b)\,\,\,\frac{{x + 1}}{{{x^2} + x + 1}}
   \end{array}$

   Giải thích các bước giải:

   \(\begin{array}{*{20}{l}}
   {a){\kern 1pt} {\kern 1pt} {\kern 1pt} \frac{1}{{x + 2}} + \frac{2}{{2 – x}} + \frac{x}{{{x^2} – 4}}}\\
   {\,\,\, = \frac{1}{{x + 2}} – \frac{2}{{x – 2}} + \frac{x}{{\left( {x – 2} \right)\left( {x + 2} \right)}}}\\
   {{\rm{}} = \frac{{x – 2 – 2\left( {x + 2} \right) + x}}{{\left( {x – 2} \right)\left( {x + 2} \right)}} = \frac{{x – 2x – 2x – 4 + x}}{{\left( {x – 2} \right)\left( {x + 2} \right)}}}\\
   {{\rm{}} = \frac{{ – 2\left( {x + 2} \right)}}{{\left( {x – 2} \right)\left( {x + 2} \right)}} = \frac{{ – 2}}{{x – 2}}.}\\
   {b){\kern 1pt} {\kern 1pt} \frac{{3{x^2} + 5x + 1}}{{{x^3} – 1}} – \frac{{1 – x}}{{{x^2} + x + 1}} – \frac{3}{{x – 1}}}\\
   {n{\rm{}} = \frac{{3{x^2} + 5x + 1}}{{\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}} + \frac{{x – 1}}{{{x^2} + x + 1}} – \frac{3}{{x – 1}}}\\
   {n{\rm{}} = \frac{{3{x^2} + 5x + 1 + {{\left( {x – 1} \right)}^2} – 3\left( {{x^2} + x + 1} \right)}}{{\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}}}\\
   {n{\rm{}} = \frac{{3{x^2} + 5x + 1 + {x^2} – 2x + 1 – 3{x^2} – 3x – 3}}{{\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}}}\\
   {n{\rm{}} = \frac{{{x^2} – 1}}{{\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}} = \frac{{\left( {x – 1} \right)\left( {x + 1} \right)}}{{\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}}}\\
   {n{\rm{}} = \frac{{x + 1}}{{{x^2} + x + 1}}.}
   \end{array}\)

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