1 câu trả lời
Đáp án: $x = - 1$
Giải thích các bước giải:
$\begin{array}{l}
Dkxd:\left\{ \begin{array}{l}
x + 1 \ge 0\\
2x + 3 \ge 0\\
x + 2 \ge 0\\
2x + 2 \ge 0
\end{array} \right. \Leftrightarrow x \ge - 1\\
\sqrt {x + 1} + \sqrt {2x + 3} = \sqrt {x + 2} + \sqrt {2x + 2} \\
\Leftrightarrow x + 1 + 2\sqrt {x + 1} .\sqrt {2x + 3} + 2x + 3\\
= x + 2 + 2\sqrt {x + 2} .\sqrt {2x + 2} + 2x + 2\\
\Leftrightarrow 2\sqrt {\left( {x + 1} \right)\left( {2x + 3} \right)} = 2\sqrt {\left( {x + 2} \right)\left( {2x + 2} \right)} \\
\Leftrightarrow \sqrt {x + 1} .\sqrt {2x + 3} .\sqrt {x + 1} .\sqrt {2\left( {x + 2} \right)} \\
\Leftrightarrow \sqrt {x + 1} .\left( {\sqrt {2x + 3} - \sqrt {2\left( {x + 2} \right)} } \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x + 1 = 0\\
2x + 3 = 2\left( {x + 2} \right)
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = - 1\left( {tm} \right)\\
2x + 3 = 2x + 4\left( {ktm} \right)
\end{array} \right.\\
Vậy\,x = - 1
\end{array}$