Đáp án:
b) \(\dfrac{3}{4}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\dfrac{1}{{x - 1}} - \dfrac{{x\left( {{x^2} - 1} \right)}}{{{x^2} + 1}}.\left( {\dfrac{1}{{{x^2} - 2x + 1}} + \dfrac{1}{{1 - {x^2}}}} \right)\\
= \dfrac{1}{{x - 1}} - \dfrac{{x\left( {{x^2} - 1} \right)}}{{{x^2} + 1}}.\left[ {\dfrac{{x + 1 - x + 1}}{{\left( {{x^2} - 1} \right)\left( {x - 1} \right)}}} \right]\\
= \dfrac{1}{{x - 1}} - \dfrac{{x\left( {{x^2} - 1} \right)}}{{{x^2} + 1}}.\dfrac{2}{{\left( {{x^2} - 1} \right)\left( {x - 1} \right)}}\\
= \dfrac{1}{{x - 1}} - \dfrac{{2x}}{{\left( {x - 1} \right)\left( {{x^2} + 1} \right)}}\\
= \dfrac{{{x^2} + 1 - 2x}}{{\left( {x - 1} \right)\left( {{x^2} + 1} \right)}}\\
= \dfrac{{{{\left( {x - 1} \right)}^2}}}{{\left( {x - 1} \right)\left( {{x^2} + 1} \right)}} = \dfrac{{x - 1}}{{{x^2} + 1}}\\
b)\left( {\dfrac{{2x - 3}}{{x{{\left( {x + 1} \right)}^2}}} + \dfrac{{4 - x}}{{x{{\left( {x + 1} \right)}^2}}}} \right):\dfrac{4}{{3x\left( {x + 1} \right)}}\\
= \dfrac{{2x - 3 + 4 - x}}{{x{{\left( {x + 1} \right)}^2}}}.\dfrac{{3x\left( {x + 1} \right)}}{4}\\
= \dfrac{{x + 1}}{{x{{\left( {x + 1} \right)}^2}}}.\dfrac{{3x\left( {x + 1} \right)}}{4}\\
= \dfrac{3}{4}
\end{array}\)
