Question

tìm x \(\sqrt{x^2-4}=8\)

Answer 1

\(\sqrt{x^2-4}=8\left(x\ge2;x\le-2\right)\\ \Leftrightarrow x^2-4=64\\ \Leftrightarrow x^2=68\\ \Leftrightarrow\left[{}\begin{matrix}x=2\sqrt{17}\left(N\right)\\x=-2\sqrt{17}\left(N\right)\end{matrix}\right.\)

Answer 2

tìm x 

<=> \(\sqrt{x^2-2^2}=8\)

\(< =>\sqrt{\left(x-2\right)^2}=8\)

\(< =>x-2=64< =>x=66\)