Cho \[{\rm{a, b}} \in {\rm{R}}\] thỏa mãn\[\mathop {{\rm{lim}}}\limits_{{\rm{x}} \to {\rm{1}}} \frac{{\sqrt {{\rm{(a + 5)}}{{\rm{x}}^{\rm{2}}} - {\rm{2(a + 2)x + 2a + b + 7}}} - \sqrt {{\rm{6x + 3}}} }}{{{{\rm{x}}^{\rm{2}}} - {\rm{2x + 1}}}}\]\[\mathop {{\rm{lim}}}\limits_{{\rm{x}} \to {\rm{1}}} \frac{{\sqrt {\left( {{\rm{a + 5}}} \right){{\rm{x}}^{\rm{2}}} - {\rm{2}}\left( {{\rm{a + 2}}} \right){\rm{x + 2a + b + 7}}} - \sqrt {{\rm{6x + 3}}} }}{{{{\rm{x}}^{\rm{2}}} - {\rm{2x + 1 }}}}{\rm{ = }}\frac{{{\rm{13}}}}{{{\rm{12}}}}\]. Tính giá trị của \[{{\rm{a}}^{\rm{2}}}{\rm{ + }}{{\rm{b}}^2}\]

Ta có \[\mathop {\lim }\limits_{{\rm{x}} \to 1} \frac{{\sqrt {{{\rm{x}}^2} + {\rm{x}} + 2} - \sqrt[3]{{7{\rm{x}} + 1}}}}{{\sqrt 2 \left( {{\rm{x}} - 1} \right)}} = \mathop {\lim }\limits_{{\rm{x}} \to 1} \frac{{\sqrt {{{\rm{x}}^2} + {\rm{x}} + 2} - 2 + 2 - \sqrt[3]{{7{\rm{x}} + 1}}}}{{\sqrt 2 \left( {{\rm{x}} - 1} \right)}}\]

\[ = \mathop {\lim }\limits_{{\rm{x}} \to 1} \frac{{\sqrt {{{\rm{x}}^2} + {\rm{x}} + 2} - 2}}{{\sqrt 2 \left( {{\rm{x}} - 1} \right)}} + \mathop {\lim }\limits_{{\rm{x}} \to 1} \frac{{2 - \sqrt[3]{{7{\rm{x}} + 1}}}}{{\sqrt 2 \left( {{\rm{x}} - 1} \right)}} = {\rm{I + J}}\]

Tính\[{\rm{I}} = \mathop {\lim }\limits_{{\rm{x}} \to 1} \frac{{\sqrt {{{\rm{x}}^2} + {\rm{x}} + 2} - 2}}{{\sqrt 2 \left( {{\rm{x}} - 1} \right)}} = \mathop {\lim }\limits_{{\rm{x}} \to 1} \frac{{{{\rm{x}}^2} + {\rm{x}} + 2 - 4}}{{\sqrt 2 \left( {{\rm{x}} - 1} \right)\left( {\sqrt {{{\rm{x}}^2} + {\rm{x}} + 2} + 2} \right)}}\]

\[ = \mathop {\lim }\limits_{{\rm{x}} \to 1} \frac{{\left( {{\rm{x}} - 1} \right)\left( {{\rm{x}} + 2} \right)}}{{\sqrt 2 \left( {{\rm{x}} - 1} \right)\left( {\sqrt {{{\rm{x}}^2} + {\rm{x}} + 2} + 2} \right)}} = \mathop {\lim }\limits_{{\rm{x}} \to 1} \frac{{{\rm{x}} + 2}}{{\sqrt 2 \left( {\sqrt {{{\rm{x}}^2} + {\rm{x}} + 2} + 2} \right)}} = \frac{3}{{4\sqrt 2 }}\]

và\[{\rm{J}} = \mathop {\lim }\limits_{{\rm{x}} \to 1} \frac{{2 - \sqrt[3]{{7{\rm{x}} + 1}}}}{{\sqrt 2 \left( {{\rm{x}} - 1} \right)}} = \mathop {\lim }\limits_{{\rm{x}} \to 1} \frac{{8 - 7{\rm{x}} - 1}}{{\sqrt 2 \left( {{\rm{x}} - 1} \right)\left[ {4 + 2\sqrt[3]{{7{\rm{x}} + 1}} + {{\left( {\sqrt[3]{{7{\rm{x}} + 1}}} \right)}^2}} \right]}}\]

\[ = \mathop {\lim }\limits_{{\rm{x}} \to 1} \frac{{ - 7}}{{\sqrt 2 \left[ {4 + 2\sqrt[3]{{7{\rm{x}} + 1}} + {{\left( {\sqrt[3]{{7{\rm{x}} + 1}}} \right)}^2}} \right]}} = \frac{{ - 7}}{{12\sqrt 2 }}\]

Do đó\[\mathop {\lim }\limits_{{\rm{x}} \to 1} \frac{{\sqrt {{{\rm{x}}^2} + {\rm{x}} + 2} - \sqrt[3]{{7{\rm{x}} + 1}}}}{{\sqrt 2 \left( {{\rm{x}} - 1} \right)}} = {\rm{I + J}} = \frac{{\sqrt 2 }}{{12}}\]

Suy ra a = 1, b = 12, c = 0. Vậy a + b + c = 13.

Đáp án cần chọn là: A

Ta có

\[\mathop {\lim }\limits_{{\rm{x}} \to {2^ + }} {\rm{f}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to {2^ + }} \left( {{{\rm{x}}^2} - 3} \right) = {2^2} - 3 = 1\]

\[\mathop {\lim }\limits_{{\rm{x}} \to {2^ - }} {\rm{f}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to {2^ - }} \left( {{\rm{x}} - 1} \right) = 2 - 1 = 1\]

\[ \Rightarrow \mathop {\lim }\limits_{{\rm{x}} \to {2^ + }} {\rm{f}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to {2^ - }} {\rm{f}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to 2} {\rm{f}}\left( {\rm{x}} \right) = 1\]

Chọn đáp án B

Đáp án cần chọn là: B