Giải thích các bước giải:
\(\begin{array}{l}
1,\\
\sqrt {{x^2} - 2x + 4} = 2x - 2\\
\Leftrightarrow \left\{ \begin{array}{l}
2x - 2 \ge 0\\
{x^2} - 2x + 4 = {\left( {2x - 2} \right)^2}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}x \ge 1\\
{x^2} - 2x + 4 = 4{x^2} - 8x + 4
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge 1\\
3{x^2} - 6x = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge 1\\
\left[ \begin{array}{l}
x = 0\\
x = 2
\end{array} \right.
\end{array} \right. \Leftrightarrow x = 2\\
2,\\
\sqrt {{x^2} - 2x} = \sqrt {2 - 3x} \,\,\,\,\,\,\,\,\,\,\,\left( {x \le 0} \right)\\
\Leftrightarrow {x^2} - 2x = 2 - 3x\\
\Leftrightarrow {x^2} + x - 2 = 0\\
\Leftrightarrow \left( {x + 2} \right)\left( {x - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = - 2\\
x = 1
\end{array} \right.\\
x \le 0 \Rightarrow x = - 2\\
3,\\
\sqrt { - {x^2} + x + 4} = x - 3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\frac{{1 - \sqrt {17} }}{2} \le x \le \frac{{1 + \sqrt {17} }}{2}} \right)\\
\Leftrightarrow \left\{ \begin{array}{l}
x - 3 \ge 0\\
- {x^2} + x + 4 = {\left( {x - 3} \right)^2}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge 3\\
- {x^2} + x + 4 = {x^2} - 6x + 9
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge 3\\
2{x^2} - 7x + 5 = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge 3\\
\left( {x - 1} \right)\left( {2x - 5} \right) = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge 3\\
\left[ \begin{array}{l}
x = 1\\
x = \frac{5}{2}
\end{array} \right.
\end{array} \right.\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {vn} \right)\\
4,\\
\sqrt {x - 3} - 2\sqrt {{x^2} - 9} = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {x \ge 3} \right)\\
\Leftrightarrow \sqrt {x - 3} - 2.\sqrt {x - 3} .\sqrt {x + 3} = 0\\
\Leftrightarrow \sqrt {x - 3} \left( {1 - 2\sqrt {x + 3} } \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt {x - 3} = 0\\
\sqrt {x + 3} = \frac{1}{2}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 3\\
x = - \frac{{11}}{4}
\end{array} \right.\\
x \ge 3 \Rightarrow x = 3\\
5,\\
2\sqrt {9x - 27} - \frac{1}{5}\sqrt {25x - 75} - \frac{1}{7}\sqrt {49x - 147} = 20\,\,\,\,\,\,\,\,\,\left( {x \ge 3} \right)\\
\Leftrightarrow 2.\sqrt {9\left( {x - 3} \right)} - \frac{1}{5}\sqrt {25\left( {x - 3} \right)} - \frac{1}{7}\sqrt {49\left( {x - 3} \right)} = 20\\
\Leftrightarrow 2.3.\sqrt {x - 3} - \frac{1}{5}.5.\sqrt {x - 3} - \frac{1}{7}.7.\sqrt {x - 3} = 20\\
\Leftrightarrow 4\sqrt {x - 3} = 20\\
\Leftrightarrow \sqrt {x - 3} = 5\\
\Leftrightarrow x = 28
\end{array}\)
