Cho a, b là các số dương. Biết \[\mathop {\lim }\limits_{{\rm{x}} \to - \infty } \left( {\sqrt {9{{\rm{x}}^2} - {\rm{ax}}} + \sqrt[3]{{{\rm{27}}{{\rm{x}}^{\rm{3}}}{\rm{ + b}}{{\rm{x}}^{\rm{2}}}{\rm{ + 5}}}}} \right) = \frac{7}{{27}}\] . Tìm giá trị lớn nhất của a. b

\[\mathop {\lim }\limits_{{\rm{x}} \to - \infty } \left( {\sqrt {9{{\rm{x}}^2} - {\rm{ax}}} + \sqrt[3]{{27{{\rm{x}}^3} + {\rm{b}}{{\rm{x}}^2} + 5}}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to - \infty } \left( {\sqrt {9{{\rm{x}}^2} - {\rm{ax}}} + {\rm{3x}} + \sqrt[3]{{27{{\rm{x}}^3} + {\rm{b}}{{\rm{x}}^2} + 5}} - 3{\rm{x}}} \right)\]

\[ = \mathop {\lim }\limits_{{\rm{x}} \to - \infty } \left( {\sqrt {9{{\rm{x}}^2} - {\rm{ax}}} + 3{\rm{x}}} \right) + \mathop {\lim }\limits_{{\rm{x}} \to - \infty } \left( {\sqrt[3]{{{\rm{27}}{{\rm{x}}^{\rm{3}}}{\rm{ + b}}{{\rm{x}}^{\rm{2}}}{\rm{ + 5}}}} - 3{\rm{x}}} \right)\]

Ta có :

\[\mathop {\lim }\limits_{{\rm{x}} \to - \infty } \left( {\sqrt {9{{\rm{x}}^2} - {\rm{ax}}} + 3{\rm{x}}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to - \infty } \frac{{ - {\rm{ax}}}}{{\sqrt {9{{\rm{x}}^2} - {\rm{ax}}} - 3{\rm{x}}}} = \mathop {\lim }\limits_{{\rm{x}} \to - \infty } \frac{{\rm{a}}}{{\sqrt {9 - \frac{{\rm{a}}}{{\rm{x}}}} + 3}} = \frac{{\rm{a}}}{{\rm{6}}}\]

Ta có :

\[\mathop {\lim }\limits_{{\rm{x}} \to - \infty } \left( {\sqrt[3]{{{\rm{27}}{{\rm{x}}^{\rm{3}}}{\rm{ + b}}{{\rm{x}}^{\rm{2}}}{\rm{ + 5}}}} - {\rm{3x}}} \right)\]

\[ = \mathop {\lim }\limits_{{\rm{x}} \to - \infty } \frac{{\left( {\sqrt[3]{{{\rm{27}}{{\rm{x}}^{\rm{3}}}{\rm{ + b}}{{\rm{x}}^{\rm{2}}}{\rm{ + 5}}}} - 3{\rm{x}}} \right)\left( {\sqrt[3]{{{{\left( {{\rm{27}}{{\rm{x}}^{\rm{3}}}{\rm{ + b}}{{\rm{x}}^{\rm{2}}}{\rm{ + 5}}} \right)}^2}}} + 3{\rm{x}}.\sqrt[3]{{{\rm{27}}{{\rm{x}}^{\rm{3}}}{\rm{ + b}}{{\rm{x}}^{\rm{2}}}{\rm{ + 5}}}} + 9{{\rm{x}}^2}} \right)}}{{\sqrt[3]{{{{\left( {{\rm{27}}{{\rm{x}}^{\rm{3}}}{\rm{ + b}}{{\rm{x}}^{\rm{2}}}{\rm{ + 5}}} \right)}^2}}} + 3{\rm{x}}.\sqrt[3]{{{\rm{27}}{{\rm{x}}^{\rm{3}}}{\rm{ + b}}{{\rm{x}}^{\rm{2}}}{\rm{ + 5}}}} + 9{{\rm{x}}^2}}}\]

\[ = \mathop {\lim }\limits_{{\rm{x}} \to - \infty } \frac{{{\rm{b}}{{\rm{x}}^{\rm{2}}}{\rm{ + 5}}}}{{\sqrt[3]{{{{\left( {{\rm{27}}{{\rm{x}}^{\rm{3}}}{\rm{ + b}}{{\rm{x}}^{\rm{2}}}{\rm{ + 5}}} \right)}^2}}} + 3{\rm{x}}.\sqrt[3]{{{\rm{27}}{{\rm{x}}^{\rm{3}}}{\rm{ + b}}{{\rm{x}}^{\rm{2}}}{\rm{ + 5}}}} + 9{{\rm{x}}^2}}}\]

\[ = \mathop {\lim }\limits_{{\rm{x}} \to - \infty } \frac{{{\rm{b + }}\frac{{\rm{5}}}{{{{\rm{x}}^{\rm{2}}}}}}}{{\sqrt[3]{{{{\left( {{\rm{27 + }}\frac{{\rm{b}}}{{\rm{x}}}{\rm{ + }}\frac{{\rm{5}}}{{{{\rm{x}}^{\rm{3}}}}}} \right)}^2}}} + 3.\sqrt[3]{{{\rm{27 + }}\frac{{\rm{b}}}{{\rm{x}}}{\rm{ + }}\frac{{\rm{5}}}{{{{\rm{x}}^{\rm{2}}}}}}} + 9}} = \frac{{\rm{b}}}{{{\rm{27}}}}\]

Do đó\[\frac{{\rm{a}}}{{\rm{6}}}{\rm{ + }}\frac{{\rm{b}}}{{{\rm{27}}}}{\rm{ = }}\frac{{\rm{7}}}{{{\rm{27}}}}\]

Áp dụng bất đẳng thức Cauchy cho 2 số dương, ta có :\[\frac{{\rm{a}}}{{\rm{6}}}{\rm{ + }}\frac{{\rm{b}}}{{{\rm{27}}}} \ge 2\sqrt {\frac{{\rm{a}}}{{\rm{6}}}{\rm{.}}\frac{{\rm{b}}}{{{\rm{27}}}}} \]

\[ \Rightarrow \frac{7}{{27}} \ge \frac{2}{{9\sqrt 2 }}\sqrt {{\rm{a}}{\rm{.b}}} \Rightarrow {\rm{ab}} \le \frac{{49}}{{18}}\]

Đẳng thức xảy ra khi\(\left\{ {\begin{array}{*{20}{c}}{\frac{{\rm{a}}}{{\rm{6}}}{\rm{ = }}\frac{{\rm{b}}}{{{\rm{2}}7}}}\\{\frac{{\rm{a}}}{{\rm{6}}} + \frac{{\rm{b}}}{{{\rm{2}}7}} = \frac{{\rm{7}}}{{{\rm{2}}7}}}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{{\rm{a}} = \frac{7}{9}}\\{{\rm{b}} = \frac{7}{2}}\end{array}} \right.\)

Vậy giá trị lớn nhất của ab bằng \[\frac{{49}}{{18}}\].

Đáp án cần chọn là: B