Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
14,\\
\dfrac{1}{6}.\sqrt {36x - 72} + \sqrt {x - 2} + 5\sqrt {x - 2} = 7\,\,\,\,\,\left( {x \ge 2} \right)\\
\Leftrightarrow \dfrac{1}{6}.\sqrt {{6^2}.\left( {x - 2} \right)} + \sqrt {x - 2} + 5\sqrt {x - 2} = 7\\
\Leftrightarrow \dfrac{1}{6}.6.\sqrt {x - 2} + \sqrt {x - 2} + 5\sqrt {x - 2} = 7\\
\Leftrightarrow \sqrt {x - 2} + \sqrt {x - 2} + 5\sqrt {x - 2} = 7\\
\Leftrightarrow 7\sqrt {x - 2} = 7\\
\Leftrightarrow \sqrt {x - 2} = 1\\
\Leftrightarrow x - 2 = 1\\
\Leftrightarrow x = 3\\
15,\\
\sqrt {16x - 16} + \sqrt {9x - 9} - \sqrt {x - 1} = 5\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {x \ge 1} \right)\\
\Leftrightarrow \sqrt {{4^2}.\left( {x - 1} \right)} + \sqrt {{3^2}.\left( {x - 1} \right)} - \sqrt {x - 1} = 5\\
\Leftrightarrow 4\sqrt {x - 1} + 3\sqrt {x - 1} - \sqrt {x - 1} = 5\\
\Leftrightarrow 6\sqrt {x - 1} = 5\\
\Leftrightarrow \sqrt {x - 1} = \dfrac{5}{6}\\
\Leftrightarrow x - 1 = \dfrac{{25}}{{36}}\\
\Leftrightarrow x = \dfrac{{61}}{{36}}\\
16,\\
\sqrt {9x + 9} - 4.\sqrt {\dfrac{{x + 1}}{4}} = 5\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {x \ge - 1} \right)\\
\Leftrightarrow \sqrt {{3^2}.\left( {x + 1} \right)} - 4.\sqrt {{{\left( {\dfrac{1}{2}} \right)}^2}.\left( {x + 1} \right)} = 5\\
\Leftrightarrow 3.\sqrt {x + 1} - 4.\dfrac{1}{2}.\sqrt {x + 1} = 5\\
\Leftrightarrow 3\sqrt {x + 1} - 2\sqrt {x + 1} = 5\\
\Leftrightarrow \sqrt {x + 1} = 5\\
\Leftrightarrow x + 1 = 25\\
\Leftrightarrow x = 24
\end{array}\)
