Question

giải phương trình:

\(\sqrt{x-2}+\sqrt{x^2-4}=0\)

Answer 1

\(ĐK:x\ge2\\ PT\Leftrightarrow\sqrt{x-2}\left(1+\sqrt{x+2}\right)=0\\ \Leftrightarrow\sqrt{x-2}=0\left(1+\sqrt{x+2}>0\right)\\ \Leftrightarrow x-2=0\Leftrightarrow x=2\left(tm\right)\)

Answer 2

ĐK:x≥2
\(< =>\sqrt{x-2}.\left(\sqrt{x+2}+1\right)=0\)
\(< =>\left[{}\begin{matrix}\sqrt{x-2}=0\\\sqrt{x+2}+1=0\end{matrix}\right.< =>\left[{}\begin{matrix}x-2=0\\\sqrt{x+2}=-1\end{matrix}\right.< =>\left[{}\begin{matrix}x=2\left(TM\right)\\x+2=1\end{matrix}\right.< =>\left[{}\begin{matrix}x=2\left(TM\right)\\x=-1\left(Loại\right)\end{matrix}\right.\)Vậy pt có nghiệm là x=2