a,$\frac{7}{x}$-$\frac{x}{x+6}$+$\frac{36}{x^2+6x}$ b, $\frac{1}{x^2-x+1}$+1-$\frac{x^2+2}{x^3+1}$

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1. Đáp án:

   $\begin{array}{l}
   a)\frac{7}{x} – \frac{x}{{x + 6}} + \frac{{36}}{{{x^2} + 6x}}\left( {Đkxđ:\left\{ \begin{array}{l}
   x \ne 0\\
   x \ne  – 6
   \end{array} \right.} \right)\\
    = \frac{7}{x} – \frac{x}{{x + 6}} + \frac{{36}}{{x\left( {x + 6} \right)}}\\
    = \frac{{7\left( {x + 6} \right) – x.x + 36}}{{x\left( {x + 6} \right)}}\\
    = \frac{{7x + 42 – {x^2} + 36}}{{x\left( {x + 6} \right)}}\\
    = \frac{{ – {x^2} + 7x + 78}}{{x\left( {x + 6} \right)}}\\
    = \frac{{ – {x^2} – 6x + 13x + 78}}{{x\left( {x + 6} \right)}}\\
    = \frac{{\left( {x + 6} \right)\left( { – x + 13} \right)}}{{x\left( {x + 6} \right)}}\\
    = \frac{{13 – x}}{x}\\
   b)\frac{1}{{{x^2} – x + 1}} + 1 – \frac{{{x^2} + 2}}{{{x^3} + 1}}\left( {Đkxđ:x \ne  – 1} \right)\\
    = \frac{1}{{{x^2} – x + 1}} + 1 – \frac{{{x^2} + 2}}{{\left( {x + 1} \right)\left( {{x^2} – x + 1} \right)}}\\
    = \frac{{x + 1 + \left( {{x^3} + 1} \right) – {x^2} – 2}}{{\left( {x + 1} \right)\left( {{x^2} – x + 1} \right)}}\\
    = \frac{{{x^3} – {x^2} + x}}{{\left( {x + 1} \right)\left( {{x^2} – x + 1} \right)}}\\
    = \frac{{x\left( {{x^2} – x + 1} \right)}}{{\left( {x + 1} \right)\left( {{x^2} – x + 1} \right)}}\\
    = \frac{x}{{x + 1}}
   \end{array}$

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