Bài 1:
a) \(\sqrt{2x+6}\) có nghĩa khi:
\(2x+6\ge0\)
\(\Leftrightarrow2x\ge-6\)
\(\Leftrightarrow x\ge-3\)
b) \(\sqrt{\dfrac{-2}{2x-3}}\) có nghĩa khi:
\(\left\{{}\begin{matrix}\dfrac{-2}{2x-3}\ge0\\2x-3\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-3\le0\left(\text{vì: }-2< 0\right)\\2x\ne3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{3}{2}\\x\ne\dfrac{3}{2}\end{matrix}\right.\)
\(\Leftrightarrow x< \dfrac{3}{2}\)
Bài 2:
a) \(\sqrt{\left(1+2\sqrt{3}\right)^2}-5\sqrt{3}\)
\(=1+2\sqrt{3}-5\sqrt{3}\)
\(=1-3\sqrt{3}\)
b) \(3\sqrt{2}+4\sqrt{8}-\sqrt{18}\)
\(=3\sqrt{2}+4\cdot2\sqrt{2}-3\sqrt{2}\)
\(=8\sqrt{2}\)
c) \(\dfrac{1}{3+\sqrt{2}}+\dfrac{1}{3-\sqrt{2}}\)
\(=\dfrac{3-\sqrt{2}}{\left(3+\sqrt{2}\right)\left(3-\sqrt{2}\right)}+\dfrac{3+\sqrt{2}}{\left(3+\sqrt{2}\right)\left(3-\sqrt{2}\right)}\)
\(=\dfrac{3-\sqrt{2}+3+\sqrt{2}}{9-2}\)
\(=\dfrac{6}{7}\)
