Tìm m để hàm số : \(f\left( x \right) = \left\{ \begin{array}{l}\frac{{\sqrt {{x^2} - 5}  - \sqrt {2x - 2} }}{{2{x^2} - 6x}}\,\,khi\

* Hướng dẫn giải

\(f(3)=2m-1\)

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to 3} f\left( x \right) = \frac{{\sqrt {{x^2} - 5}  - \sqrt {2x - 2} }}{{2{x^2} - 6x}} = \mathop {\lim }\limits_{x \to 3} \frac{{{x^2} - 5 - 2x + 2}}{{2x\left( {x - 3} \right)\left( {\sqrt {{x^2} - 5}  + \sqrt {2x - 2} } \right)}}\\
\mathop {\lim }\limits_{x \to 3} \frac{{\left( {x - 3} \right)\left( {x + 1} \right)}}{{2x\left( {x - 3} \right)\left( {\sqrt {{x^2} - 5}  + \sqrt {2x - 2} } \right)}} = \mathop {\lim }\limits_{x \to 3} \frac{{x + 1}}{{2x\left( {\sqrt {{x^2} - 5}  + \sqrt {2x - 2} } \right)}} = \frac{1}{6}
\end{array}\)

Hàm số liên tục tại \({x_0} = 3 \Leftrightarrow f\left( 3 \right) = \mathop {\lim }\limits_{x \to 3} f\left( x \right) \Rightarrow 2m - 1 = \frac{1}{6} \Leftrightarrow m = \frac{7}{{12}}\)