ĐK: \(\sqrt{x-2m}-3\ne0\Leftrightarrow x-2m\ne9\Leftrightarrow x\ne9+2m\)

Hàm số xác đinh trên khoảng (3; 5) 

<=>  2m + 9 \(\le\)3 hoặc 2m + 9 \(\ge\)5

<=> m \(\le\)-3 hoặc m \(\ge\)-2

ĐKXĐ: \(\left\{{}\begin{matrix}x-m+1\ge0\\-x+2m>0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge m-1\\x< 2m\end{matrix}\right.\)

\(\Rightarrow x\in[m-1;2m)\)

Để hàm xác định trên (3;4)

\(\Rightarrow\left(3;4\right)\subset[m-1;2m)\)

\(\Rightarrow\left\{{}\begin{matrix}m-1\le3\\2m\ge4\end{matrix}\right.\) \(\Rightarrow2\le m\le4\)

\(a,\)\(đkxđ\Leftrightarrow x\ge0\)

\(b,\)\(A=\left(\frac{1}{\sqrt{x}+1}-\frac{1}{x+\sqrt{x}}\right):\frac{x-\sqrt{x}+1}{x\sqrt{x}+1}.\)

\(=\left(\frac{1}{\sqrt{x}+1}-\frac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right):\frac{x-\sqrt{x}+1}{\sqrt{x}^3+1}\)

\(=\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}+1\right)}:\frac{x-\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)

\(=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}-1}{\sqrt{x}}\)

\(c,\)\(A\ge0\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}}\ge0\)

Mà \(\sqrt{x}\ge0\Rightarrow\sqrt{x}-1\ge0\Rightarrow\sqrt{x}\ge1\Rightarrow x\ge1\)