Đáp án đúng:
Giải chi tiết:Lời giải:
a) ĐK:
\(\left\{ \begin{array}{l}{x^3} - 1 \ne 0\\x - 1 \ne 0\\\frac{x}{{{x^2} + x + 1}} - \frac{2}{{x - 1}} \ne 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ne 1\\x \ne - 2\end{array} \right..\)
\(\begin{array}{l}A = \frac{{\frac{{{x^4} + 1}}{{{x^3} - 1}} - x}}{{\frac{x}{{{x^2} + x + 1}} - \frac{2}{{x - 1}}}}\\\,\,\,\,\, = \frac{{{x^4} + 1 - {x^4} + x}}{{{x^3} - 1}}:\frac{{x(x - 1) - 2({x^2} + x + 1)}}{{{x^3} - 1}}\\\,\,\,\,\,\, = \frac{{1 + x}}{{{x^3} - 1}}.\frac{{{x^3} - 1}}{{{x^2} - x - 2{x^2} - 2x - 2}}\\\,\,\,\,\, = \frac{{x + 1}}{{ - {x^2} - 3x - 2}}\\\,\,\,\,\,\, = - \frac{{x + 1}}{{(x + 1)(x + 2)}}\\\,\,\,\,\,\, = - \frac{1}{{x + 2}}.\end{array}\)
b) ĐK:
\(\left\{ \begin{array}{l}{x^3} + {x^2} + x \ne 0\\{x^4} - x \ne 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ne 0\\x \ne 1\end{array} \right..\)
\(\begin{array}{l}B = \frac{{{x^2} + 1}}{{{x^3} + {x^2} + x}}:\frac{1}{{{x^4} - x}}\\\,\,\,\,\, = \frac{{{x^2} + 1}}{{x({x^2} + x + 1)}}.x({x^3} - 1)\\\,\,\,\,\, = \frac{{{x^2} + 1}}{{x({x^2} + x + 1)}}.\frac{{x(x - 1)({x^2} + x + 1)}}{1}\\\,\,\,\,\, = (x - 1)({x^2} + 1).\end{array}\)
