Cho hàm số \(f\left( x \right) = \left\{ \begin{array}{l}\frac{{{x^2} - 6x + 5}}{{{x^2} - 1}}\,\,khi\,\,x \ne 1\\a + \frac{5}{2}\,\,\,\,\,\,\,\,\

* Hướng dẫn giải

TXĐ: D = R

\(\mathop {\lim }\limits_{x \to 1} \frac{{{x^2} - 6x + 5}}{{{x^2} - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {x - 1} \right)\left( {x - 5} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} = \mathop {\lim }\limits_{x \to 1} \frac{{x - 5}}{{x + 1}} =  - 2\)

\(f\left( 1 \right) = a + \frac{5}{2}\)

Hàm số liên tục tại \(x = 1 \Leftrightarrow \mathop {\lim }\limits_{x \to 1} f\left( x \right) = f\left( 1 \right) \Leftrightarrow a + \frac{5}{2} =  - 2 \Leftrightarrow a =  - \frac{9}{2}\).