Đáp án:
a) \(x \ne \pm 1,\,\,\,P = \dfrac{{x - 7}}{{x + 1}}\).
b) \(x = 0 \Rightarrow P = - 7\).
Giải thích các bước giải:
\(P = \dfrac{x}{{x - 1}} + \dfrac{3}{{x + 1}} - \dfrac{{6x - 4}}{{{x^2} - 1}}\)
a) ĐKXĐ: \(\left\{ \begin{array}{l}x - 1 \ne 0\\x + 1 \ne 0\\{x^2} - 1 \ne 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ne 1\\x \ne - 1\\x \ne \pm 1\end{array} \right. \Leftrightarrow x \ne \pm 1\).
\(\begin{array}{l}P = \dfrac{x}{{x - 1}} + \dfrac{3}{{x + 1}} - \dfrac{{6x - 4}}{{{x^2} - 1}}\\P = \dfrac{{x\left( {x + 1} \right) - 3\left( {x - 1} \right) - 6x + 4}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\\P = \dfrac{{{x^2} + x - 3x + 3 - 6x + 4}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\\P = \dfrac{{{x^2} - 8x + 7}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\\P = \dfrac{{{x^2} - x - 7x + 7}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\\P = \dfrac{{x\left( {x - 1} \right) - 7\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\\P = \dfrac{{\left( {x - 1} \right)\left( {x - 7} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\\P = \dfrac{{x - 7}}{{x + 1}}\end{array}\)
b) Với \({x^2} - x = 0 \Leftrightarrow x\left( {x - 1} \right) = 0 \Leftrightarrow x = 0\,\,\left( {tmDK} \right)\) hoặc \(x = 1\) (ktm ĐK)
Với \(x = 0\) (tm ĐK) thì \(P = \dfrac{{0 - 7}}{{0 + 1}} = \dfrac{{ - 7}}{1} = - 7\).
