Question

Answer 1

\(a,ĐK:\dfrac{2}{x-3}\ge0\Leftrightarrow x-3>0\left(2>0;x-3\ne0\right)\Leftrightarrow x>3\\ b,ĐK:x\ge0;y\ge0;\sqrt{x}-\sqrt{y}\ne0\Leftrightarrow x\ne y\)

Answer 2

a) Để \(A=\sqrt{\dfrac{2}{x-3}}\) có nghĩa thì:

\(\dfrac{2}{x-3}\ge0\)\(\Leftrightarrow\left\{{}\begin{matrix}x-3\ge0\\x-3\ne0\end{matrix}\right.\)\(\Leftrightarrow x>3\)

b) Để \(B=\dfrac{1}{\sqrt{x}-\sqrt{y}}\) có nghĩa thì:

\(\left\{{}\begin{matrix}x\ge0,y\ge0\\x\ne y\end{matrix}\right.\)