Đáp án:

\(\begin{array}{l}
1,\\
\left\{ \begin{array}{l}
x \ne 1\\
x \ne 0\\
x \ne  - 1
\end{array} \right.\\
2,\\
Phương\,\,trình\,\,vô\,\,nghiệm
\end{array}\)

Giải thích các bước giải:

\(\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^2} + x \ne 0\\
{x^3} + x \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{x^2} + x + 1 \ne 0\\
{x^3} \ne 1\\
x \ne 1\\
x\left( {x + 1} \right) \ne 0\\
x\left( {{x^2} + 1} \right) \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ne 1\\
x \ne 0\\
x \ne  - 1
\end{array} \right.\\
2,\\
R = \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}}\\
 = \left[ {\dfrac{{{{\left( {x - 1} \right)}^2}}}{{3x + {x^2} - 2.x.1 + {1^2}}} - \dfrac{{1 - 2{x^2} + 4x}}{{{x^3} - {1^3}}} + \dfrac{1}{{x - 1}}} \right]:\dfrac{{x.\left( {x + 1} \right)}}{{x\left( {{x^2} + 1} \right)}}\\
 = \left[ {\dfrac{{{{\left( {x - 1} \right)}^2}}}{{3x + {x^2} - 2x + 1}} - \dfrac{{1 - 2{x^2} + 4x}}{{\left( {x - 1} \right)\left( {{x^2} + x.1 + {1^2}} \right)}} + \dfrac{1}{{x - 1}}} \right]:\dfrac{{x + 1}}{{{x^2} + 1}}\\
 = \left[ {\dfrac{{{{\left( {x - 1} \right)}^2}}}{{{x^2} + x + 1}} - \dfrac{{1 - 2{x^2} + 4x}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} + \dfrac{1}{{x - 1}}} \right]:\dfrac{{x + 1}}{{{x^2} + 1}}\\
 = \dfrac{{{{\left( {x - 1} \right)}^3} - \left( {1 - 2{x^2} + 4x} \right) + \left( {{x^2} + x + 1} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}:\dfrac{{x + 1}}{{{x^2} + 1}}\\
 = \dfrac{{\left( {{x^3} - 3.{x^2}.1 + 3.x{{.1}^2} - {1^3}} \right) - 1 + 2{x^2} - 4x + {x^2} + x + 1}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}:\dfrac{{x + 1}}{{{x^2} + 1}}\\
 = \dfrac{{{x^3} - 3{x^2} + 3x - 1 + 3{x^2} - 3x}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}.\dfrac{{{x^2} + 1}}{{x + 1}}\\
 = \dfrac{{{x^3} - 1}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}.\dfrac{{{x^2} + 1}}{{x + 1}}\\
 = 1.\dfrac{{{x^2} + 1}}{{x + 1}}\\
 = \dfrac{{{x^2} + 1}}{{x + 1}}\\
R = 0 \Leftrightarrow \dfrac{{{x^2} + 1}}{{x + 1}} = 0 \Leftrightarrow {x^2} + 1 = 0 \Leftrightarrow {x^2} =  - 1\\
 \Rightarrow Phương\,\,trình\,\,vô\,\,nghiệm
\end{array}\)