Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\sqrt {3x - 4} = x\,\,\,\,\,\,\,\,\,\left( {x \ge \dfrac{4}{3}} \right)\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
3x - 4 = {x^2}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
{x^2} - 3x + 4 = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
\left( {{x^2} - 3x + \dfrac{9}{4}} \right) + \dfrac{3}{4} = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
{\left( {x - \dfrac{3}{2}} \right)^2} + \dfrac{3}{4} = 0\,\,\,\,\,\left( {vn} \right)
\end{array} \right.\\
\Rightarrow ptvn\\
b,\\
\sqrt {{x^2} - 2x + 2} = 2x - 1\\
\Leftrightarrow \left\{ \begin{array}{l}
2x - 1 \ge 0\\
{x^2} - 2x + 2 = {\left( {2x - 1} \right)^2}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge \dfrac{1}{2}\\
{x^2} - 2x + 2 = 4{x^2} - 4x + 1
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge \dfrac{1}{2}\\
3{x^2} - 2x - 1 = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge \dfrac{1}{2}\\
\left( {x - 1} \right)\left( {3x + 1} \right) = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge \dfrac{1}{2}\\
\left[ \begin{array}{l}
x = 1\\
x = - \dfrac{1}{3}
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow x = 1
\end{array}\)
