Đáp án:
\(\begin{array}{l}
a,\\
x = - 1\\
b,\\
\left[ \begin{array}{l}
x = 4\\
x = 12
\end{array} \right.\\
c,\\
x = 25\\
d,\\
x = \dfrac{3}{2}
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
DKXD:\,\,\,x \ge - 5\\
\sqrt {4x + 20} - 2\sqrt {x + 5} + \sqrt {9x + 45} = 6\\
\Leftrightarrow \sqrt {4\left( {x + 5} \right)} - 2\sqrt {x + 5} + \sqrt {9\left( {x + 5} \right)} = 6\\
\Leftrightarrow \sqrt {{2^2}\left( {x + 5} \right)} - 2\sqrt {x + 5} + \sqrt {{3^2}.\left( {x + 5} \right)} = 6\\
\Leftrightarrow 2\sqrt {x + 5} - 2\sqrt {x + 5} + 3\sqrt {x + 5} = 6\\
\Leftrightarrow 3\sqrt {x + 5} = 6\\
\Leftrightarrow \sqrt {x + 5} = 2\\
\Leftrightarrow x + 5 = {2^2}\\
\Leftrightarrow x = - 1\\
b,\\
DKXD:\,\,\,\,\,x \ge 4\\
\sqrt {{x^2} - 16} - 4\sqrt {x - 4} = 0\\
\Leftrightarrow \sqrt {\left( {x - 4} \right)\left( {x + 4} \right)} - 4\sqrt {x - 4} = 0\\
\Leftrightarrow \sqrt {x - 4} .\left( {\sqrt {x + 4} - 4} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt {x - 4} = 0\\
\sqrt {x + 4} - 4 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\sqrt {x - 4} = 0\\
\sqrt {x + 4} = 4
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x - 4 = 0\\
x + 4 = 16
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 4\\
x = 12
\end{array} \right.\\
c,\\
DKXD:\,\,\,x \ge 0\\
3\sqrt x = 15\\
\Leftrightarrow \sqrt x = 5\\
\Leftrightarrow x = {5^2}\\
\Leftrightarrow x = 25\\
d,\\
DKXD:\,\,\,\,x \ge - \dfrac{1}{2}\\
\sqrt {2x + 1} = 2\\
\Leftrightarrow 2x + 1 = {2^2}\\
\Leftrightarrow 2x + 1 = 4\\
\Leftrightarrow 2x = 3\\
\Leftrightarrow x = \dfrac{3}{2}
\end{array}\)
