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Câu hỏi: Cho $\lim \dfrac{\left( a{{n}^{2}}-n \right)\left( 2n-1 \right)}{\left( 1+b{{n}^{2}} \right)\left( 5-3n \right)}=3$, với $a, b\ne 0$. Khẳng định nào sau đây đúng
A. $a=-\dfrac{9b}{2}$.
B. $b=-9a$.
C. $a=9b$.
D. $b=-3a$.
Mà $\lim \dfrac{\left( a{{n}^{2}}-n \right)\left( 2n-1 \right)}{\left( 1+b{{n}^{2}} \right)\left( 5-3n \right)}=3$ $\Rightarrow \dfrac{2a}{-3b}=3\Leftrightarrow a=-\dfrac{9b}{2}$.
A. $a=-\dfrac{9b}{2}$.
B. $b=-9a$.
C. $a=9b$.
D. $b=-3a$.
Ta có $\lim \dfrac{\left( a{{n}^{2}}-n \right)\left( 2n-1 \right)}{\left( 1+b{{n}^{2}} \right)\left( 5-3n \right)}=\lim \dfrac{\dfrac{\left( a{{n}^{2}}-n \right)\left( 2n-1 \right)}{{{n}^{3}}}}{\dfrac{\left( 1+b{{n}^{2}} \right)\left( 5-3n \right)}{{{n}^{3}}}}=\lim \dfrac{\left( a-\dfrac{1}{n} \right)\left( 2-\dfrac{1}{n} \right)}{\left( \dfrac{1}{{{n}^{2}}}+b \right)\left( \dfrac{5}{n}-3 \right)}=\dfrac{2a}{-3b}$.Mà $\lim \dfrac{\left( a{{n}^{2}}-n \right)\left( 2n-1 \right)}{\left( 1+b{{n}^{2}} \right)\left( 5-3n \right)}=3$ $\Rightarrow \dfrac{2a}{-3b}=3\Leftrightarrow a=-\dfrac{9b}{2}$.
Đáp án A.