Giải thích các bước giải:
\(\begin{array}{l}
a,\\
{\rm{DK:}}\,\,\,x > \frac{3}{4}\\
\frac{{{x^2}}}{{\sqrt {4x - 3} }} - \sqrt {4x - 3} = 1 - x\\
\Leftrightarrow {x^2} - \left( {4x - 3} \right) = \left( {1 - x} \right).\sqrt {4x - 3} \\
\Leftrightarrow {x^2} - 4x + 3 - \left( {1 - x} \right)\sqrt {4x - 3} = 0\\
\Leftrightarrow \left( {x - 1} \right)\left( {x - 3} \right) + \left( {x - 1} \right)\sqrt {4x - 3} = 0\\
\Leftrightarrow \left( {x - 1} \right)\left[ {\left( {x - 3} \right) + \sqrt {4x - 3} } \right] = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 1 = 0\\
\left( {x - 3} \right) + \sqrt {4x - 3} = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
\sqrt {4x - 3} = 3 - x
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
\left\{ \begin{array}{l}
3 - x \ge 0\\
4x - 3 = {\left( {3 - x} \right)^2}
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
\left\{ \begin{array}{l}
x \le 3\\
4x - 3 = {x^2} - 6x + 9
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
\left\{ \begin{array}{l}
x \le 3\\
{x^2} - 10x + 12 = 0
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
\left\{ \begin{array}{l}
x \le 3\\
x = 5 \pm \sqrt {13}
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = 5 - \sqrt {13}
\end{array} \right.\\
b,\\
{\rm{DK:}}\,\,\,\,\,x \ge \frac{1}{2}\\
4{x^2} + 3x + 3 = 4x\sqrt {x + 3} + 2\sqrt {2x - 1} \\
\Leftrightarrow 4{x^2} + 3x + 3 - 4x.\sqrt {x + 3} - 2\sqrt {2x - 1} = 0\\
\Leftrightarrow \left[ {4{x^2} - 4x.\sqrt {x + 3} + \left( {x + 3} \right)} \right] + \left[ {\left( {2x - 1} \right) - 2\sqrt {2x - 1} + 1} \right] = 0\\
\Leftrightarrow {\left( {2x - \sqrt {x + 3} } \right)^2} + {\left( {\sqrt {2x - 1} - 1} \right)^2} = 0\\
\Leftrightarrow \left\{ \begin{array}{l}
2x - \sqrt {x + 3} = 0\\
\sqrt {2x - 1} - 1 = 0
\end{array} \right. \Leftrightarrow x = 1
\end{array}\)
