Question

1. tính giới hạn sau:

\(\lim\sqrt{2n^{2}-4}-7n\)

2. tìm a,b để:

\(a^{2}-2; 8; a; 2b \) là 1 CSC

Answer 1

\(lim\sqrt{2n^2-4}-7n\)

\(=lim\sqrt{n^2\left(2-\dfrac{4}{n^2}\right)}-7n=n\sqrt{2}-7n=-\infty\)

Answer 2

2:

Theo đề, ta có: \(\left\{{}\begin{matrix}a^2-2+a=16\\a=\dfrac{2b+8}{2}=b+4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a^2+a-18=0\\a=b+4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}a\in\left\{\dfrac{-1+\sqrt{73}}{2};\dfrac{-1-\sqrt{73}}{2}\right\}\\b\in\left\{\dfrac{7+\sqrt{73}}{2};\dfrac{7-\sqrt{73}}{2}\right\}\end{matrix}\right.\)