Đáp án:
c) \(x < 1;x \ne 0\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ne \left\{ {0;1} \right\}\\
P = \dfrac{{{x^2} + x}}{{{x^2} - x + 1}}:\left[ {\dfrac{{x + 1}}{x} - \dfrac{1}{{1 - x}} + \dfrac{{2 - {x^2}}}{{{x^2} - x}}} \right]\\
= \dfrac{{x\left( {x + 1} \right)}}{{{x^2} - x + 1}}:\left[ {\dfrac{{\left( {x + 1} \right)\left( {x - 1} \right) + x + 2 - {x^2}}}{{x\left( {x - 1} \right)}}} \right]\\
= \dfrac{{x\left( {x + 1} \right)}}{{{x^2} - x + 1}}.\dfrac{{x\left( {x - 1} \right)}}{{{x^2} - 1 + x + 2 - {x^2}}}\\
= \dfrac{{x\left( {x + 1} \right)}}{{{x^2} - x + 1}}.\dfrac{{x\left( {x - 1} \right)}}{{x + 1}}\\
= \dfrac{{{x^2}\left( {x - 1} \right)}}{{{x^2} - x + 1}}\\
c)P < 0\\
\to \dfrac{{{x^2}\left( {x - 1} \right)}}{{{x^2} - x + 1}} < 0\\
Do:\left\{ \begin{array}{l}
{x^2} > 0\forall x \ne \left\{ {0;1} \right\}\\
{x^2} - x + 1 > 0\forall x \ne \left\{ {0;1} \right\}
\end{array} \right.\\
\to P < 0\\
\Leftrightarrow x - 1 < 0\\
\to x < 1;x \ne 0
\end{array}\)
