Đáp án:
c) x>1
Giải thích các bước giải:
\(\begin{array}{l}
B1:\\
a)DK:x \ne \pm 1\\
b)H = \left( {\dfrac{1}{{x - 1}} - \dfrac{{2x}}{{x\left( {{x^2} + 1} \right) - \left( {{x^2} + 1} \right)}}} \right):\left( {\dfrac{{{x^2} + 1 - 2x}}{{{x^2} + 1}}} \right)\\
= \left( {\dfrac{1}{{x - 1}} - \dfrac{{2x}}{{\left( {{x^2} + 1} \right)\left( {x - 1} \right)}}} \right):\dfrac{{{{\left( {x - 1} \right)}^2}}}{{{x^2} + 1}}\\
= \dfrac{{{x^2} + 1 - 2x}}{{\left( {{x^2} + 1} \right)\left( {x - 1} \right)}}.\dfrac{{{x^2} + 1}}{{{{\left( {x - 1} \right)}^2}}}\\
= \dfrac{{{{\left( {x - 1} \right)}^2}}}{{\left( {{x^2} + 1} \right)\left( {x - 1} \right)}}.\dfrac{{{x^2} + 1}}{{{{\left( {x - 1} \right)}^2}}}\\
= \dfrac{1}{{x - 1}}\\
c)H > 0\\
\to \dfrac{1}{{x - 1}} > 0\\
\to x - 1 > 0\\
\to x > 1
\end{array}\)
