Question

Giải phương trình : \(\sqrt{x^2-7x+10}+6042=3\sqrt{x-2}+2014\sqrt{x-5}\)

Answer 1

ĐKXĐ: \(x\ge5\)

\(\Leftrightarrow\sqrt{\left(x-2\right)\left(x-5\right)}+6042-3\sqrt{x-2}-2014\sqrt{x-5}=0\)

\(\Leftrightarrow\sqrt{x-5}\left(\sqrt{x-2}-2014\right)-3\left(\sqrt{x-2}-2014\right)=0\)

\(\Leftrightarrow\left(\sqrt{x-5}-3\right)\left(\sqrt{x-2}-2014\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-5}=3\\\sqrt{x-2}=2014\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=9\\x-2=2014^2\end{matrix}\right.\)