đk \(\left\{{}\begin{matrix}x-2m+3\ge0\\x\ne m\\-x+m+5\ge0\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}x\ge2m-3\\x\ne m\\x\le m+5\end{matrix}\right.\)
=> TXĐ:D=[2m-3;m+5]\{m}
Để hàm số xác định trên khoảng (0;1) thì (0;1) là con của D=[2m-3;m+5]\{m}
<=>\(\left\{{}\begin{matrix}2m-3\le0\\m+5\ge1\\\left[{}\begin{matrix}m\le0\\m\ge1\end{matrix}\right.\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}m\le\dfrac{3}{2}\\m\ge-4\\\left[{}\begin{matrix}m\le0\\m\ge1\end{matrix}\right.\end{matrix}\right.\)
TH1
\(\left\{{}\begin{matrix}m\le\dfrac{3}{2}\\m\ge-4\\m\le0\end{matrix}\right.\)
.
<=>\(\left\{{}\begin{matrix}m\le0\\m\ge-4\end{matrix}\right.\)
<=> m thuộc [-4;0]
Th2
\(\left\{{}\begin{matrix}m\le\dfrac{3}{2}\\m\ge-4\\m\ge1\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}m\ge1\\m\le\dfrac{3}{2}\end{matrix}\right.\)
<=> m thuộc [1;\(\dfrac{3}{2}\)]
