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Câu hỏi: Cho các số thực a, b, c thỏa mãn ${{c}^{2}}+a=18$ và $\underset{x\to +\infty }{\mathop{\lim }} \left( \sqrt{a{{x}^{2}}+b\text{x}}-c\text{x} \right)=-2.$ Tính $P=a+b+5c$
A. $P=18$
B. $P=12$
C. $P=9$
D. $P=5$
A. $P=18$
B. $P=12$
C. $P=9$
D. $P=5$
Dựa vào giả thiết suy ra $\left\{ \begin{aligned}
& a>0 \\
& c>0 \\
\end{aligned} \right.$
Ta có: $\underset{x\to +\infty }{\mathop{\lim }} \left( \sqrt{a{{x}^{2}}+b\text{x}}-c\text{x} \right)=-2\Leftrightarrow \underset{x\to +\infty }{\mathop{\lim }} \dfrac{a{{x}^{2}}+b\text{x}-{{c}^{2}}{{\text{x}}^{2}}}{\sqrt{a{{x}^{2}}+b\text{x}}+c\text{x}}=-2$
$\Leftrightarrow \underset{x\to +\infty }{\mathop{\lim }} \dfrac{\left( a-{{c}^{2}} \right){{\text{x}}^{2}}+b\text{x}}{\sqrt{a{{x}^{2}}+b\text{x}}+c\text{x}}=-2\Leftrightarrow \left\{ \begin{aligned}
& a={{c}^{2}} \\
& \dfrac{b}{\sqrt{a}+c}=-2 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=9;c=3 \\
& b=-12 \\
\end{aligned} \right.\Rightarrow $ $ P=a+b+5c=12$
& a>0 \\
& c>0 \\
\end{aligned} \right.$
Ta có: $\underset{x\to +\infty }{\mathop{\lim }} \left( \sqrt{a{{x}^{2}}+b\text{x}}-c\text{x} \right)=-2\Leftrightarrow \underset{x\to +\infty }{\mathop{\lim }} \dfrac{a{{x}^{2}}+b\text{x}-{{c}^{2}}{{\text{x}}^{2}}}{\sqrt{a{{x}^{2}}+b\text{x}}+c\text{x}}=-2$
$\Leftrightarrow \underset{x\to +\infty }{\mathop{\lim }} \dfrac{\left( a-{{c}^{2}} \right){{\text{x}}^{2}}+b\text{x}}{\sqrt{a{{x}^{2}}+b\text{x}}+c\text{x}}=-2\Leftrightarrow \left\{ \begin{aligned}
& a={{c}^{2}} \\
& \dfrac{b}{\sqrt{a}+c}=-2 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=9;c=3 \\
& b=-12 \\
\end{aligned} \right.\Rightarrow $ $ P=a+b+5c=12$
Đáp án B.