Cho a và b là các số nguyên dương. Biết đường thẳng y = 7/27

* Hướng dẫn giải

Đáp án B

$$\begin{gathered} \underset{x\to -\infty }{\lim }y=\underset{x\to -\infty }{\lim }\left(\sqrt{9x^2+ax}+\sqrt[3]{27x^3+b{x}^2+5}\right) \\ =\underset{x\to -\infty }{\lim }\left[\left(\sqrt{9x^2+ax}+3x\right)+\left(\sqrt[3]{27x^3+b{x}^2+5}-3x\right)\right] \\ =\underset{x\to -\infty }{\lim }\left[\frac{9x^2+ax-9x^2}{\sqrt{9x^2+ax}-3x}+\frac{27x^3+b{x}^2+5-27x^3}{\sqrt[3]{{\left(27x^3+b{x}^2+5\right)}^2}+3x\sqrt[3]{27x^3+b{x}^2+5}+9x^2}\right] \\ =\underset{x\to -\infty }{\lim }\left(\frac{ax}{\sqrt{9x^2+ax}-3x}+\frac{b{x}^2+5}{\sqrt[3]{{\left(27x^3+b{x}^2+5\right)}^2}+3x\sqrt[3]{27x^3+b{x}^2+5}+9x^2}\right) \\ =\underset{x\to -\infty }{\lim }\left(\frac{a}{-\sqrt{9+\frac{a}{x}}-3}+\frac{b+\frac{5}{x^2}}{\sqrt[3]{{\left(27+\frac{b}{x}+\frac{5}{x^3}\right)}^2}+3\sqrt[3]{27+\frac{b}{x}+\frac{5}{x^3}}+9}\right) \\ =\frac{{\displaystyle a}}{{\displaystyle -3-3}}+\frac{{\displaystyle b}}{{\displaystyle 9+3.3+9}}=\frac{{\displaystyle a}}{{\displaystyle -6}}+\frac{{\displaystyle b}}{{\displaystyle 27}}=\frac{{\displaystyle 7}}{{\displaystyle 27}}\Rightarrow -\frac{{\displaystyle 9}}{{\displaystyle 2}}.\frac{{\displaystyle a}}{{\displaystyle 27}}+\frac{{\displaystyle b}}{{\displaystyle 27}}=7 \\ \Rightarrow -9a+2b=14 \end{gathered}$$