Giải pt Hộ e vs mn e c.ơn

Đáp án:

\(\dfrac{1}{{x + 1}}\)

Giải thích các bước giải:

\(\begin{array}{l}
\dfrac{1}{{x – 1}} – \dfrac{{{x^3} – x}}{{{x^2} + x}}.\left( {\dfrac{1}{{{x^2} – 2x + 1}} + \dfrac{1}{{1 – {x^2}}}} \right)\\
 = \dfrac{1}{{x – 1}} – \dfrac{{x\left( {x – 1} \right)\left( {x + 1} \right)}}{{x\left( {x + 1} \right)}}.\left[ {\dfrac{1}{{{{\left( {x – 1} \right)}^2}}} – \dfrac{1}{{\left( {x – 1} \right)\left( {x + 1} \right)}}} \right]\\
 = \dfrac{1}{{x – 1}} – \dfrac{{x\left( {x – 1} \right)\left( {x + 1} \right)}}{{x\left( {x + 1} \right)}}.\left[ {\dfrac{{x + 1 – x + 1}}{{\left( {x – 1} \right)\left( {x + 1} \right)\left( {x – 1} \right)}}} \right]\\
 = \dfrac{1}{{x – 1}} – \left( {x – 1} \right).\dfrac{2}{{\left( {x – 1} \right)\left( {x + 1} \right)\left( {x – 1} \right)}}\\
 = \dfrac{1}{{x – 1}} – \dfrac{2}{{\left( {x – 1} \right)\left( {x + 1} \right)}}\\
 = \dfrac{{x + 1 – 2}}{{\left( {x – 1} \right)\left( {x + 1} \right)}}\\
 = \dfrac{{x – 1}}{{\left( {x – 1} \right)\left( {x + 1} \right)}}\\
 = \dfrac{1}{{x + 1}}
\end{array}\)