Question

biết \(\lim\limits_{x\rightarrow-\infty}\left(\sqrt{2x^2-3x 1} x\sqrt{2}\right)=\dfrac{a}{b}\sqrt{2}\)  ( a nguyen, b nguyen duong, a/b toi gian). tìm a, b?

Answer 1

\(\lim\limits_{x\rightarrow-\infty}\dfrac{2x^2-\left(2x^2-3x+1\right)}{x\sqrt{2}-\sqrt{2x^2-3x+1}}=\lim\limits_{x\rightarrow-\infty}\dfrac{3x-1}{x\sqrt{2}-\sqrt{2x^2-3x+1}}\)

\(=\lim\limits_{x\rightarrow-\infty}\dfrac{3-\dfrac{1}{x}}{\sqrt{2}+\sqrt{2-\dfrac{3}{x}+\dfrac{1}{x^2}}}=\dfrac{3}{2\sqrt{2}}=\dfrac{3}{4}\sqrt{2}\)

\(\Rightarrow\left\{{}\begin{matrix}a=3\\b=4\end{matrix}\right.\)

Answer 2

\(\lim\limits_{x\rightarrow+\infty}\left(\sqrt{2x^2-3x+1}+x\sqrt{2}\right)=+\infty\) nên chắc chắn đề bài sai

Đề đúng sẽ là: \(x\rightarrow-\infty\) hoặc \(x\rightarrow+\infty\) thì biểu thức là \(\sqrt{2x^2-3x+1}-x\sqrt{2}\)