Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
c,\\
\left( {{x^2} - 1} \right)\left( {x + 2} \right)\left( {x - 3} \right) = \left( {x - 1} \right)\left( {{x^2} - 4} \right)\left( {x + 5} \right)\\
\Leftrightarrow \left( {x - 1} \right)\left( {x + 1} \right)\left( {x + 2} \right)\left( {x - 3} \right) = \left( {x - 1} \right)\left( {x - 2} \right)\left( {x + 2} \right)\left( {x + 5} \right)\\
\Leftrightarrow \left( {x - 1} \right).\left( {x + 2} \right).\left[ {\left( {x + 1} \right)\left( {x - 3} \right)} \right] = \left( {x - 1} \right)\left( {x + 2} \right).\left[ {\left( {x - 2} \right).\left( {x + 5} \right)} \right]\\
\Leftrightarrow \left[ \begin{array}{l}
x - 1 = 0\\
x + 2 = 0\\
\left( {x + 1} \right)\left( {x - 3} \right) = \left( {x - 2} \right)\left( {x + 5} \right)
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = - 2\\
{x^2} - 3x + x - 3 = {x^2} + 5x - 2x - 10
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = - 2\\
{x^2} - 2x - 3 = {x^2} + 3x - 10
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = - 2\\
- 2x - 3 = 3x - 10
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = - 2\\
x = \dfrac{7}{5}
\end{array} \right.\\
\Rightarrow x \in \left\{ {1; - 2;\dfrac{7}{5}} \right\}\\
d,\\
{x^4} + {x^3} + x + 1 = 0\\
\Leftrightarrow \left( {{x^4} + {x^3}} \right) + \left( {x + 1} \right) = 0\\
\Leftrightarrow {x^3}.\left( {x + 1} \right) + \left( {x + 1} \right) = 0\\
\Leftrightarrow \left( {x + 1} \right).\left( {{x^3} + 1} \right) = 0\\
\Leftrightarrow \left( {x + 1} \right).\left( {x + 1} \right)\left( {{x^2} - x + 1} \right) = 0\\
\Leftrightarrow {\left( {x + 1} \right)^2}\left( {{x^2} - x + 1} \right) = 0\\
{x^2} - x + 1 = \left( {{x^2} - x + \dfrac{1}{4}} \right) + \dfrac{3}{4} = {\left( {x - \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} \ge \dfrac{3}{4} > 0,\,\,\,\forall x\\
\Rightarrow {\left( {x + 1} \right)^2} = 0\\
\Leftrightarrow x + 1 = 0\\
\Leftrightarrow x = - 1
\end{array}\)
