Rút gọn biểu thức \(P\left( x \right) = \left( {\dfrac{1}{{{x^2} - x}} + \dfrac{1}{{x - 1}}} \right):\dfrac{{x + 1}}{{{x^2} - 2x + 1}}.\)

* Hướng dẫn giải

Ta có \(P\left( x \right)\) xác định \( \Leftrightarrow \left\{ \begin{array}{l}{x^2} - x \ne 0\\x - 1 \ne 0\\x + 1 \ne 0\\{x^2} - 2x + 1 \ne 0\end{array} \right. \)

\(\Leftrightarrow \left\{ \begin{array}{l}x\left( {x - 1} \right) \ne 0\\x \ne  \pm 1\\{\left( {x - 1} \right)^2} \ne 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ne 0\\x \ne  \pm 1\end{array} \right..\)

\(\begin{array}{l}P\left( x \right) = \left( {\dfrac{1}{{{x^2} - x}} + \dfrac{1}{{x - 1}}} \right):\dfrac{{x + 1}}{{{x^2} - 2x + 1}}\\\;\;\;\;\;\;\;\; = \left( {\dfrac{1}{{x\left( {x - 1} \right)}} + \dfrac{1}{{x - 1}}} \right):\dfrac{{x + 1}}{{{{\left( {x - 1} \right)}^2}}}\\\;\;\;\;\;\;\;\; = \dfrac{{x + 1}}{{x\left( {x - 1} \right)}}.\dfrac{{{{\left( {x - 1} \right)}^2}}}{{x + 1}}\\\;\;\;\;\;\;\;\; = \dfrac{{x - 1}}{x}.\end{array}\)