Đáp án:
\(\begin{array}{l}
1,\\
\left[ \begin{array}{l}
x = 2\\
x = 1
\end{array} \right.\\
2,\\
x = 2\\
3,\\
\left[ \begin{array}{l}
x = - 3\\
x = 12
\end{array} \right.
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
DKXD:\,\,\,x \ge - 3\\
\sqrt {{x^2} + 10x + 21} = 3\sqrt {x + 3} + 2\sqrt {x + 7} - 6\\
\Leftrightarrow \sqrt {{x^2} + 10x + 21} - 3\sqrt {x + 3} - 2\sqrt {x + 7} + 6 = 0\\
\Leftrightarrow \sqrt {\left( {x + 3} \right)\left( {x + 7} \right)} - 3\sqrt {x + 3} - 2\sqrt {x + 7} + 6 = 0\\
\Leftrightarrow \left( {\sqrt {\left( {x + 3} \right)\left( {x + 7} \right)} - 3\sqrt {x + 3} } \right) + \left( { - 2\sqrt {x + 7} + 6} \right) = 0\\
\Leftrightarrow \sqrt {x + 3} \left( {\sqrt {x + 7} - 3} \right) - 2\left( {\sqrt {x + 7} - 3} \right) = 0\\
\Leftrightarrow \left( {\sqrt {x + 7} - 3} \right)\left( {\sqrt {x + 3} - 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt {x + 7} - 3 = 0\\
\sqrt {x + 3} - 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\sqrt {x + 7} = 3\\
\sqrt {x + 3} = 2
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x + 7 = 9\\
x + 3 = 4
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x = 1
\end{array} \right.\\
2,\\
DKXD:\,\,\,x \ge 2\\
\sqrt {{x^2} - 4} - 2\sqrt {x - 2} = 0\\
\Leftrightarrow \sqrt {\left( {x - 2} \right)\left( {x + 2} \right)} - 2\sqrt {x - 2} = 0\\
\Leftrightarrow \sqrt {x - 2} \left( {\sqrt {x + 2} - 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt {x - 2} = 0\\
\sqrt {x + 2} - 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\sqrt {x - 2} = 0\\
\sqrt {x + 2} = 2
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x - 2 = 0\\
x + 2 = 4
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x = 2
\end{array} \right. \Leftrightarrow x = 2\\
3,\\
DKXD:\,\,\,\,\left[ \begin{array}{l}
x \ge 3\\
x = - 3
\end{array} \right.\\
\sqrt {{x^2} - 9} - 3\sqrt {x + 3} = 0\\
\Leftrightarrow \sqrt {\left( {x - 3} \right)\left( {x + 3} \right)} - 3\sqrt {x + 3} = 0\\
\Leftrightarrow \sqrt {x + 3} \left( {\sqrt {x - 3} - 3} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt {x + 3} = 0\\
\sqrt {x - 3} - 3 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\sqrt {x + 3} = 0\\
\sqrt {x - 3} = 3
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x + 3 = 0\\
x - 3 = 9
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = - 3\\
x = 12
\end{array} \right.
\end{array}\)
