Em ghi rõ đề bài ra nhé.
Giải thích các bước giải:
\(\begin{array}{l}
\sqrt {2\left( {x + 1} \right)} \,\,DK:\,x + 1 \ge 0 \Leftrightarrow x \ge - 1\\
\sqrt {x + \dfrac{1}{2}\sqrt x + \dfrac{1}{2}} \\
DK:\,\left\{ \begin{array}{l}
x \ge 0\\
x + \dfrac{1}{2}\sqrt x + \dfrac{1}{2} \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
{\left( {x + \dfrac{1}{4}} \right)^2} + \dfrac{3}{{16}} > 0\left( {ld} \right)
\end{array} \right.\\
\Rightarrow x \ge 0\\
\sqrt {2x + \dfrac{1}{3}\sqrt {2x} + \dfrac{1}{3}} \,\,DK:\,x \ge 0\\
\sqrt x + \sqrt {x + 1} \,\,DK:\,\left\{ \begin{array}{l}
x \ge 0\\
x + 1 \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
x \ge - 1
\end{array} \right. \Leftrightarrow x \ge 0\\
\sqrt {x + 3} + \sqrt {x - 5} \\
DK:\left\{ \begin{array}{l}
x + 3 \ge 0\\
x - 5 \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge - 3\\
x \ge 5
\end{array} \right. \Leftrightarrow x \ge 5
\end{array}\)
