Đáp án:
\[x = 2\]
Giải thích các bước giải:
ĐKXĐ: \(x \ge 1\)
Ta có:
\(\begin{array}{l}
{x^2} - 3x + 4 = 2\sqrt {x - 1} \\
\Leftrightarrow {x^2} - 3x + 4 - 2\sqrt {x - 1} = 0\\
\Leftrightarrow \left( {{x^2} - 4x + 4} \right) + \left( {x - 2\sqrt {x - 1} } \right) = 0\\
\Leftrightarrow {\left( {x - 2} \right)^2} + \left[ {\left( {x - 1} \right) - 2.\sqrt {x - 1} .1 + 1} \right] = 0\\
\Leftrightarrow {\left( {x - 2} \right)^2} + {\left( {\sqrt {x - 1} - 1} \right)^2} = 0\\
{\left( {x - 2} \right)^2} \ge 0,\,\,\,\forall x\\
{\left( {\sqrt {x - 1} - 1} \right)^2} \ge 0,\,\,\,\forall x \ge 1\\
\Rightarrow {\left( {x - 2} \right)^2} + {\left( {\sqrt {x - 1} - 1} \right)^2} \ge 0,\,\,\,\forall x \ge 1\\
\Rightarrow \left\{ \begin{array}{l}
{\left( {x - 2} \right)^2} = 0\\
{\left( {\sqrt {x - 1} - 1} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 2\\
\sqrt {x - 1} = 1
\end{array} \right. \Leftrightarrow x = 2\,\,\,\,\left( {t/m} \right)
\end{array}\)
Vậy \(x = 2\)
