Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
*)\\
\dfrac{{\sqrt {2 - 4x} }}{{\sqrt {x - 2} }} = 2\\
DKXD:\,\,\,\left\{ \begin{array}{l}
2 - 4x \ge 0\\
x - 2 > 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
4x \le 2\\
x > 2
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \le \dfrac{1}{2}\\
x > 2
\end{array} \right.\\
\Rightarrow phương\,\,\,\,trình\,\,\,\,vô\,\,\,\,nghiệm\\
*)\\
1 + \sqrt {3x + 1} = 3x\\
DKXD:\,\,\,3x + 1 \ge 0 \Leftrightarrow x \ge - \dfrac{1}{3}\\
1 + \sqrt {3x + 1} = 3x\\
\Leftrightarrow \sqrt {3x + 1} = 3x - 1\\
\Leftrightarrow \left\{ \begin{array}{l}
3x - 1 \ge 0\\
{\sqrt {3x + 1} ^2} = {\left( {3x - 1} \right)^2}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge \dfrac{1}{3}\\
3x + 1 = 9{x^2} - 6x + 1
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge \dfrac{1}{3}\\
9{x^2} - 9x = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge \dfrac{1}{3}\\
9x\left( {x - 1} \right) = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge \dfrac{1}{3}\\
\left[ \begin{array}{l}
x = 0\\
x = 1
\end{array} \right.
\end{array} \right. \Leftrightarrow x = 1\,\,\,\,\left( {t/m} \right)\\
b,\\
*)\\
\sqrt {\dfrac{{2 - 4x}}{{x - 2}}} = 2\\
DKXD:\,\,\,\,\left\{ \begin{array}{l}
\dfrac{{2 - 4x}}{{x - 2}} \ge 0\\
x - 2 \ne 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
2 - 4x \ge 0\\
x - 2 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
2 - 4x \le 0\\
x - 2 < 0
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \le \dfrac{1}{2}\\
x > 2
\end{array} \right.\\
\left\{ \begin{array}{l}
x \ge \dfrac{1}{2}\\
x < 2
\end{array} \right.
\end{array} \right. \Leftrightarrow \dfrac{1}{2} \le x < 2\\
\sqrt {\dfrac{{2 - 4x}}{{x - 2}}} = 2\\
\Leftrightarrow \dfrac{{2 - 4x}}{{x - 2}} = 4\\
\Leftrightarrow 2 - 4x = 4.\left( {x - 2} \right)\\
\Leftrightarrow 2 - 4x = 4x - 8\\
\Leftrightarrow 10 = 8x\\
\Leftrightarrow x = \dfrac{5}{4}\,\,\,\,\left( {t/m} \right)\\
*)\\
\sqrt {5{x^2} + 1} = 2x\\
DKXD:\,\,\,5{x^2} + 1 \ge 0,\,\,\,\forall x\\
\sqrt {5{x^2} + 1} = 2x\\
\Leftrightarrow 5{x^2} + 1 = {\left( {2x} \right)^2}\\
\Leftrightarrow 5{x^2} + 1 = 4{x^2}\\
\Leftrightarrow {x^2} = - 1\\
\Rightarrow phương\,\,\,trình\,\,\,vô\,\,\,\,nghiệm\\
c,\\
*)\\
\sqrt {3 + \sqrt x } = 3\\
DKXD:\,\,\,x \ge 0\\
\sqrt {3 + \sqrt x } = 3\\
\Leftrightarrow 3 + \sqrt x = 9\\
\Leftrightarrow \sqrt x = 6\\
\Leftrightarrow x = 36\,\,\,\,\left( {t/m} \right)\\
*)\\
\sqrt {x - \sqrt {2x - 1} } = 2\\
DKXD:\,\,\,\left\{ \begin{array}{l}
2x - 1 \ge 0\\
x - \sqrt {2x - 1} \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge \dfrac{1}{2}\\
x \ge \sqrt {2x - 1}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge \dfrac{1}{2}\\
x \ge 0\\
{x^2} \ge 2x - 1
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge \dfrac{1}{2}\\
{x^2} - 2x + 1 \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge \dfrac{1}{2}\\
{\left( {x - 1} \right)^2} \ge 0
\end{array} \right. \Rightarrow x \ge \dfrac{1}{2}\\
\sqrt {x - \sqrt {2x - 1} } = 2\\
\Leftrightarrow x - \sqrt {2x - 1} = 4\\
\Leftrightarrow \sqrt {2x - 1} = x - 4\\
\Leftrightarrow \left\{ \begin{array}{l}
x - 4 \ge 0\\
{\sqrt {2x - 1} ^2} = {\left( {x - 4} \right)^2}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge 4\\
2x - 1 = {x^2} - 8x + 16
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge 4\\
{x^2} - 10x + 17 = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge 4\\
x = 5 \pm 2\sqrt 2
\end{array} \right. \Rightarrow x = 5 + 2\sqrt 2 \,\,\,\left( {t/m} \right)\\
d,\\
3\sqrt {5 - 2x} > 27\\
DKXD:\,\,\,\,5 - 2x \ge 0 \Leftrightarrow x \le \dfrac{5}{2}\\
3\sqrt {5 - 2x} > 27\\
\Leftrightarrow \sqrt {5 - 2x} > 9\\
\Leftrightarrow 5 - 2x > 81\\
\Leftrightarrow 2x < - 76\\
\Leftrightarrow x < - 38\,\,\,\,\,\left( {t/m} \right)
\end{array}\)
