Chứng minh rằng nếu $ab \neq-4$ thì hàm số: $y = \frac{2x+a}{bx-2}$ trùng với hàm số ngược của nó.
* Khi $b = 0  \Rightarrow y = \frac{{2{\rm{x}} + a}}{{ - 2}} \Leftrightarrow x = \frac{{2y + a}}{{ - 2}}$
Vậy $y$ có hàm số ngược trùng với chính nó.
* Khi $b \neq 0 : y = \frac{{2{\rm{x}} + a}}{{b{\rm{x}} - 2}} = \frac{2}{b} + \frac{{ab +
4}}{{b(bx - 2)}}, x \in R   $ \ $  \left\{ {\frac{2}{5}} \right\}$
Nếu $ab + 4 = 0 \Leftrightarrow ab = -4  \Rightarrow y = \frac{2}{b}$ là hằng số, nên không có hàm số ngược.
* Khi $ab\neq- 4 : y = \frac{{2x + a}}{{bx - 2}} \Leftrightarrow x = \frac{{2y + a}}{{by - 2}}, y \in R $ \ $ \left\{ {\frac{2}{5}} \right\}$
Ta nhận thấy $2$ hàm số $y = \frac{{2x + a}}{{bx - 2}}$ và $x = \frac{{2y + a}}{{by - 2}}$
có cùng miền xác định $D = R $ \ $ \left\{ {\frac{2}{5}} \right\}$ trùng nhau.

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