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2 câu trả lời 6163
\[\begin{array}{l}
\sqrt {2{x^2} - 3x + 1} = x - 1\\
dk:2{x^2} - 3x + 1 \ge 0\\
\Leftrightarrow \left[ \begin{array}{l}
x \ge 1\\
x \le \frac{1}{2}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x - 1 \ge 0\\
2{{\rm{x}}^2} - 3{\rm{x}} + 1 = {\left( {x - 1} \right)^2}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge 1\\
2{{\rm{x}}^2} - 3{\rm{x}} + 1 = {x^2} - 2{\rm{x}} + 1
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge 1\\
{x^2} - x = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge 1\\
x\left( {x - 1} \right) = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge 1\\
\left[ \begin{array}{l}
x = 0\\
x - 1 = 0
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge 1\\
\left[ \begin{array}{l}
x = 0\\
x = 1
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow x = 1\left( {tm} \right)\\
vay:x = 1
\end{array}\]
2x^2-3x+1=(x-1)^2
-> 2x^2-3x+1=x^2-2x+1
-> x=0 (ktm) và x=1 (tm)
-> x=1
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