Đáp án:
x=2
Giải thích các bước giải:
$\begin{array}{l}
\sqrt {x - 1} + \sqrt {{x^3} + {x^2} + x + 1} = 1 + \sqrt {{x^4} - 1} \,\left( {dkxd\,\left\{ \begin{array}{l}
x - 1 \ge 0\\
x + 1 \ge 0
\end{array} \right. \Rightarrow x \ge 1\,} \right)\\
\Leftrightarrow \sqrt {x - 1} + \sqrt {{x^3} + {x^2} + x + 1} = 1 + \sqrt {\left( {x - 1} \right)\left( {{x^3} + {x^2} + x + 1} \right)} \\
\Leftrightarrow \sqrt {x - 1} + \sqrt {{x^3} + {x^2} + x + 1} - 1 - \sqrt {x - 1} .\sqrt {{x^3} + {x^2} + x + 1} = 0\\
\Leftrightarrow \left( {\sqrt {x - 1} - 1} \right)\left( {\sqrt {{x^3} + {x^2} + x + 1} - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt {x - 1} = 1\\
\sqrt {{x^3} + {x^2} + x + 1} = 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x - 1 = 1\\
{x^3} + {x^2} + x + 1 = 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 2\\
{x^3} + {x^2} + x = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 2\left( {tmdk} \right)\\
x = 0\left( {ktmdk} \right)
\end{array} \right.
\end{array}$
