\(x-13\sqrt{x}+40=0\Leftrightarrow x-8\sqrt{x}-5\sqrt{x}+40=0\)
\(\Leftrightarrow\left(x-8\sqrt{x}\right)-\left(5\sqrt{x}-40\right)=0\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-8\right)-5\left(\sqrt{x}-8\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-8\right)\left(\sqrt{x}-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-8=0\\\sqrt{x}-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=8\\\sqrt{x}=5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\pm2\sqrt{2}\\x=\pm\sqrt{5}\end{matrix}\right.\)
giải:
\(x-13\sqrt{x}+40=0\) ĐK: x≥0
\(\Leftrightarrow x-5\sqrt{x}-8\sqrt{x}+40=0\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-5\right)-8\left(\sqrt{x}-5\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-8\right)\left(\sqrt{x}-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-8=0\\\sqrt{x}-5=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=8\\\sqrt{x}=25\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=64\left(tm\right)\\x=25\left(tm\right)\end{matrix}\right.\)
Vậy x ∈ \(\left\{64;25\right\}\)
