Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\sqrt {{x^2} - 4x + 4} = \sqrt {4{x^2} - 12x + 9} \\
\Leftrightarrow \sqrt {{{\left( {x - 2} \right)}^2}} = \sqrt {{{\left( {2x - 3} \right)}^2}} \\
\Leftrightarrow \left| {x - 2} \right| = \left| {2x - 3} \right|\\
\Leftrightarrow \left[ \begin{array}{l}
x - 2 = 2x - 3\\
x - 2 = 3 - 2x
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = \frac{5}{3}
\end{array} \right.\\
b,\\
\sqrt {x + 2\sqrt {x - 1} } = 2\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {x \ge 1} \right)\\
\Leftrightarrow \sqrt {\left( {x - 1} \right) + 2\sqrt {x - 1} + 1} = 2\\
\Leftrightarrow \sqrt {{{\left( {\sqrt {x - 1} + 1} \right)}^2}} = 2\\
\Leftrightarrow \sqrt {x - 1} + 1 = 2\\
\Leftrightarrow \sqrt {x - 1} = 1\\
\Leftrightarrow x = 2\\
c,\\
\sqrt {{x^2} - 2x + 1} = 9x - 1\\
\Leftrightarrow \sqrt {{{\left( {x - 1} \right)}^2}} = 9x - 1\\
\Leftrightarrow \left| {x - 1} \right| = 9x - 1\\
\Leftrightarrow \left\{ \begin{array}{l}
9x - 1 \ge 0\\
\left[ \begin{array}{l}
x - 1 = 9x - 1\\
x - 1 = 1 - 9x
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge \frac{1}{9}\\
\left[ \begin{array}{l}
x = 0\\
x = \frac{1}{5}
\end{array} \right.
\end{array} \right. \Leftrightarrow x = \frac{1}{5}\\
d,\\
\sqrt {{x^2} - 6x + 9} = 4 - x\\
\Leftrightarrow \sqrt {{{\left( {x - 3} \right)}^2}} = 4 - x\\
\Leftrightarrow \left| {x - 3} \right| = 4 - x\\
\Leftrightarrow \left\{ \begin{array}{l}
4 - x \ge 0\\
\left[ \begin{array}{l}
x - 3 = 4 - x\\
x - 3 = x - 3
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \le 4\\
x = \frac{7}{2}
\end{array} \right. \Leftrightarrow x = \frac{7}{2}\\
e,\\
\sqrt {{x^2} - 9} + \sqrt {{x^2} - 6x + 9} = 0\\
\Leftrightarrow \left\{ \begin{array}{l}
{x^2} - 9 = 0\\
{x^2} - 6x + 9 = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left( {x - 3} \right)\left( {x + 3} \right) = 0\\
{\left( {x - 3} \right)^2} = 0
\end{array} \right.\\
\Leftrightarrow x = 3
\end{array}\)
