Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\sqrt {3x - 6} = 4\,\,\,\,\,\,\,\,\,\,\,\left( {x \ge 2} \right)\\
\Leftrightarrow {\sqrt {3x - 6} ^2} = {4^2}\\
\Leftrightarrow 3x - 6 = 16\\
\Leftrightarrow 3x = 16 + 6\\
\Leftrightarrow 3x = 22\\
\Leftrightarrow x = \frac{{22}}{3}\\
b,\\
\sqrt {{{\left( {4x - 7} \right)}^2}} = 3\\
\Leftrightarrow \left| {4x - 7} \right| = 3\\
\Leftrightarrow \left[ \begin{array}{l}
4x - 7 = 3\\
4x - 7 = - 3
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
4x = 10\\
4x = 4
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \frac{5}{2}\\
x = 1
\end{array} \right.\\
c,\\
\sqrt {{x^2} + 4x + 4} = 6\\
\Leftrightarrow \sqrt {{{\left( {x + 2} \right)}^2}} = 6\\
\Leftrightarrow \left| {x + 2} \right| = 6\\
\Leftrightarrow \left[ \begin{array}{l}
x + 2 = 6\\
x + 2 = - 6
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 4\\
x = - 8
\end{array} \right.\\
d,\\
\sqrt {8a} - \sqrt {18a} + \sqrt {2a} - 4 = 0\\
\Leftrightarrow \sqrt {4.2a} - \sqrt {9.2a} + \sqrt {2a} - 4 = 0\\
\Leftrightarrow \sqrt {{2^2}.2a} - \sqrt {{3^2}.2a} + \sqrt {2a} - 4 = 0\\
\Leftrightarrow 2\sqrt {2a} - 3\sqrt {2a} + \sqrt {2a} - 4 = 0\\
\Leftrightarrow - 4 = 0\,\,\,\,\,\,\left( {ptvn} \right)
\end{array}\)
Đề câu e bị thiếu rồi em
