Question

(x+3)\(\sqrt{x^2-4}\)\(=x^2-9\)

Answer 1

ĐKXĐ: \(\left\{{}\begin{matrix}x\ge2\\x\le-2\end{matrix}\right.\)

\(\left(x+3\right)\sqrt{x^2-4}=\left(x+3\right)\left(x-3\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\\sqrt{x^2-4}=x-3\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}x\ge3\\x^2-4=x^2-6x+9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge3\\x=\dfrac{13}{6}< 3\left(loại\right)\end{matrix}\right.\)

Vậy pt có nghiệm duy nhất \(x=-3\)