Cho a và b là các số nguyên dương. Biết đường thẳng

* Hướng dẫn giải

Đáp án B

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

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

$=\underset{x\to -\infty }{\lim }( \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} )$

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

$=\underset{x\to -\infty }{\lim }\left(\frac{a}{-\sqrt{9+\frac{a}{x}}-3}+\frac{b+\frac{5}{x^2}}{\sqrt[3]{{\left(27+\frac{b}{x}+\frac{5}{x^3}\right)}^2}+3\sqrt[3]{27+\frac{b}{x}+\frac{5}{x^3}}+9}\right)$

$\begin{array}{l}=\frac{a}{-3-3}+\frac{b}{9+3.3+9}\\ =\frac{a}{-6}+\frac{b}{27}=\frac{7}{27}\\ \Rightarrow -\frac{9}{2}.\frac{a}{27}+\frac{b}{27}=7\\ \Rightarrow -9a+2b=14\end{array}$