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Câu hỏi: Đồ thị hàm số $y=\dfrac{{{x}^{3}}-4x}{{{x}^{3}}-3x-2}$ có bao nhiêu đường tiệm cận?
A. 4.
B. 1.
C. 3.
D. 2.
A. 4.
B. 1.
C. 3.
D. 2.
$y=\dfrac{x\left( x-2 \right)\left( x+2 \right)}{\left( x+1 \right)\left( {{x}^{2}}-x-2 \right)}=\dfrac{x\left( x-2 \right)\left( x+2 \right)}{{{\left( x+1 \right)}^{2}}\left( x-2 \right)}=\dfrac{x\left( x+2 \right)}{{{\left( x+1 \right)}^{2}}}\Rightarrow TC\tilde{N}:x=-1.$
Đồng thời $\left\{ \begin{aligned}
& \underset{x\to +\infty }{\mathop{\lim }} y=\underset{x\to +\infty }{\mathop{\lim }} \dfrac{1-\dfrac{4}{{{x}^{2}}}}{1-\dfrac{3}{{{x}^{2}}}-\dfrac{2}{{{x}^{3}}}}=1\Rightarrow TCN:y=1 \\
& \underset{x\to -\infty }{\mathop{\lim }} y=\underset{x\to -\infty }{\mathop{\lim }} \dfrac{1-\dfrac{4}{{{x}^{2}}}}{1-\dfrac{3}{{{x}^{2}}}-\dfrac{2}{{{x}^{3}}}}=1\Rightarrow TCN:y=1 \\
\end{aligned} \right.\Rightarrow $
Đồng thời $\left\{ \begin{aligned}
& \underset{x\to +\infty }{\mathop{\lim }} y=\underset{x\to +\infty }{\mathop{\lim }} \dfrac{1-\dfrac{4}{{{x}^{2}}}}{1-\dfrac{3}{{{x}^{2}}}-\dfrac{2}{{{x}^{3}}}}=1\Rightarrow TCN:y=1 \\
& \underset{x\to -\infty }{\mathop{\lim }} y=\underset{x\to -\infty }{\mathop{\lim }} \dfrac{1-\dfrac{4}{{{x}^{2}}}}{1-\dfrac{3}{{{x}^{2}}}-\dfrac{2}{{{x}^{3}}}}=1\Rightarrow TCN:y=1 \\
\end{aligned} \right.\Rightarrow $
Đáp án D.