Đáp án:
$\begin{array}{l}
1)Dkxd:\left\{ \begin{array}{l}
3x + {\left( {x - 1} \right)^2} \ne 0\\
{x^3} - 1 \ne 0\\
x - 1 \ne 0\\
{x^3} + x \ne 0\\
{x^2} + x \ne 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{x^2} - 2x + 1 + 3x \ne 0\\
x \ne 1\\
x \ne 1\\
x\left( {{x^2} + 1} \right) \ne 0\\
x\left( {x + 1} \right) \ne 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{x^2} + x + 1 \ne 0\left( {tm} \right)\\
x \ne 1\\
x \ne 0\\
x \ne - 1
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ne 0\\
x \ne 1\\
x \ne - 1
\end{array} \right.\\
Vậy\,Dkxd:x \ne 0;x \ne 1;x \ne - 1\\
2)R = 0\\
\Leftrightarrow \left[ {\dfrac{{{{\left( {x - 1} \right)}^2}}}{{3x + {{\left( {x - 1} \right)}^2}}} - \dfrac{{1 - 2{x^2} + 4x}}{{{x^3} - 1}} + \dfrac{1}{{x - 1}}} \right]:\dfrac{{{x^2} + x}}{{{x^3} + x}} = 0\\
\Leftrightarrow \dfrac{{{{\left( {x - 1} \right)}^2}}}{{3x + {{\left( {x - 1} \right)}^2}}} - \dfrac{{1 - 2{x^2} + 4x}}{{{x^3} - 1}} + \dfrac{1}{{x - 1}} = 0\\
\Leftrightarrow \dfrac{{{{\left( {x - 1} \right)}^2}}}{{{x^2} + x + 1}} + \dfrac{{2{x^2} - 4x - 1}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} + \dfrac{1}{{x - 1}} = 0\\
\Leftrightarrow \dfrac{{{{\left( {x - 1} \right)}^2}\left( {x - 1} \right) + 2{x^2} - 4x - 1 + {x^2} + x + 1}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} = 0\\
\Leftrightarrow {\left( {x - 1} \right)^3} + 3{x^2} - 3x = 0\\
\Leftrightarrow {x^3} - 3{x^2} + 3x - 1 + 3{x^2} - 3x = 0\\
\Leftrightarrow {x^3} = 1\\
\Leftrightarrow x = 1\left( {ktm} \right)
\end{array}$
Vậy ko có x thỏa mãn để R=0
