ĐKXĐ:\(x\ge5\)
\(\sqrt{x^2-25}-2\sqrt{x-5}=0\)
\(\Leftrightarrow\sqrt{\left(x-5\right)\left(x+5\right)}-2\sqrt{x-5}=0\)
\(\Leftrightarrow\sqrt{x-5}\left(\sqrt{x+5}-2\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-5}=0\\\sqrt{x+5}-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-5=0\\\sqrt{x+5}=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)
vì x=-1 không thỏa mãn được điều kiện vì thế x=5
\(\sqrt{\left(x-5\right)\left(x+5\right)}-2\sqrt{x-5}\)=0
\(\sqrt{x-5}\sqrt{x+5}-2\sqrt{x-5}\)=0
\(\sqrt{x-5}\left(\sqrt{x+5}-2\right)\)=0
\(\left[{}\begin{matrix}\sqrt{x-5}=0\\\sqrt{x+5}-2=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=5\\x+5=4\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)
