Đáp án đúng:
Giải chi tiết:Lời giải:
a) ĐK:
\(\left\{ \begin{array}{l}3x + {\left( {x - 1} \right)^2} \ne 0\\{x^3} - 1 \ne 0\\{x^3} + x \ne 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ne \pm 1\\x \ne 0\end{array} \right..\)
\(\begin{array}{l}Q = \left[ {\frac{{{{(x - 1)}^2}}}{{3x + {{(x - 1)}^2}}} - \frac{{1 - 2{x^2} + 4x}}{{{x^3} - 1}} + \frac{1}{{x - 1}}} \right]:\frac{{3x}}{{{x^3} + x}}\\\,\,\,\,\, = \left[ {\frac{{{{\left( {x - 1} \right)}^2}}}{{3x + {x^2} - 2x + 1}} - \frac{{1 - 2{x^2} + 4x}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} + \frac{1}{{x - 1}}} \right]:\frac{{3x}}{{x\left( {{x^2} + 1} \right)}}\\\,\,\,\,\, = \left[ {\frac{{{{\left( {x - 1} \right)}^2}}}{{{x^2} + x + 1}} + \frac{{2{x^2} - 4x - 1}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} + \frac{1}{{x - 1}}} \right].\frac{{x({x^2} + 1)}}{{3x}}\\\,\,\,\,\, = \frac{{{{(x - 1)}^3} + 2{x^2} - 4x - 1 + {x^2} + x + 1}}{{{x^3} - 1}}.\frac{{{x^2} + 1}}{3}\\\,\,\,\,\, = \frac{{{x^3} - 3x{}^2 + 3x - 1 + 2{x^2} - 4x - 1 + {x^2} + x + 1}}{{{x^3} - 1}}.\frac{{{x^2} + 1}}{3}\\\,\,\,\,\, = \frac{{{x^3} - 1}}{{{x^3} - 1}}.\frac{{{x^2} + 1}}{3} = \frac{{{x^2} + 1}}{3}.\end{array}\)
b) Ta có Q = \(\frac{{{x}^{2}}+1}{3}\),
\({{x}^{2}}\ge 0\,\,\forall x\Rightarrow {{x}^{2}}+1\ge 1\,\,\forall x\Rightarrow \frac{{{x}^{2}}+1}{3}\ge \frac{1}{3}\forall x\)
Vậy \(Max\,\,Q=\frac{1}{3}.\)
