Question

Tính giá trị hàm số

\(a^2+ab-b^2\)

\(\lim\limits_{x\rightarrow-\infty}\frac{x\sqrt{x^2+2x}+x^2}{\sqrt{16x^2+1}-x+2}=\frac{a}{b}\)

Answer 1

\(\lim\limits_{x\rightarrow-\infty}\frac{x\left(\sqrt{x^2+2x}+x\right)\left(\sqrt{x^2+2x}-x\right)}{\left(\sqrt{16x^2+1}-x+2\right)\left(\sqrt{x^2+2x}-x\right)}\)

\(=\lim\limits_{x\rightarrow-\infty}\frac{2x^2}{\left(\sqrt{16x^2+1}-x+2\right)\left(\sqrt{x^2+2x}-x\right)}=\frac{2}{\left(-\sqrt{16+\frac{1}{x^2}}-1+\frac{2}{x}\right)\left(-\sqrt{1+\frac{2}{x}}-1\right)}=\frac{2}{-5.-2}=\frac{1}{5}\)

\(\Rightarrow\left\{{}\begin{matrix}a=1\\b=5\end{matrix}\right.\) \(\Rightarrow a^2+ab-b^2=-20\)