Đáp án đúng: D
Phương pháp giải:
Hàm số \(y = f\left( x \right)\) liên tục tại \(x = {x_0}\) khi và chỉ khi \(\mathop {\lim }\limits_{x \to {x_0}} f\left( x \right) = f\left( {{x_0}} \right)\).
Giải chi tiết:TXĐ: \(D = \left[ {\dfrac{1}{2}; + \infty } \right)\) và \(x = 1 \in D\).
\(\begin{array}{l}\mathop {\lim }\limits_{x \to 1} f\left( x \right) = \mathop {\lim }\limits_{x \to 1} \dfrac{{\sqrt {2x - 1} - \sqrt[3]{{3{x^2} - 3x + 1}}}}{{{{\left( {x - 1} \right)}^2}}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to 1} \dfrac{{\sqrt {2x - 1} - x}}{{{{\left( {x - 1} \right)}^2}}} + \mathop {\lim }\limits_{x \to 1} \dfrac{{x - \sqrt[3]{{3{x^2} - 3x + 1}}}}{{{{\left( {x - 1} \right)}^2}}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to 1} \dfrac{{2x - 1 - {x^2}}}{{{{\left( {x - 1} \right)}^2}\left( {\sqrt {2x - 1} + x} \right)}} + \mathop {\lim }\limits_{x \to 1} \dfrac{{{x^3} - 3{x^2} + 3x - 1}}{{{{\left( {x - 1} \right)}^2}\left[ {{x^2} + x.\sqrt[3]{{3{x^2} - 3x + 1}} + {{\sqrt[3]{{3{x^2} - 3x + 1}}}^2}} \right]}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to 1} \dfrac{{ - 1}}{{\sqrt {2x - 1} + x}} + \mathop {\lim }\limits_{x \to 1} \dfrac{{x - 1}}{{{x^2} + x.\sqrt[3]{{3{x^2} - 3x + 1}} + {{\sqrt[3]{{3{x^2} - 3x + 1}}}^2}}} = \dfrac{{ - 1}}{2}\\f\left( 1 \right) = {m^2} + 2m - \dfrac{3}{2}\end{array}\)
Để hàm số đã cho liên tục tại \(x = 1\) thì \(\mathop {\lim }\limits_{x \to 1} f\left( x \right) = f\left( 1 \right)\) \( \Leftrightarrow {m^2} + 2m - \dfrac{3}{2} = - \dfrac{1}{2} \Leftrightarrow m = - 1 \pm \sqrt 2 \).
Vậy \(m = - 1 \pm \sqrt 2 \).