ĐKXĐ : \(x\ge0\)
\(\Leftrightarrow\) \(-\left(5x-7\sqrt{x}-12\right)=0\)
\(\Leftrightarrow\) \(-\left(5x+5\sqrt{x}-12\sqrt{x}-12\right)=0\)
\(\Leftrightarrow\) \(-\left[5\sqrt{x}\left(\sqrt{x}+1\right)-12\left(\sqrt{x}+1\right)\right]=0\)
\(\Leftrightarrow\) \(-\left(5\sqrt{x}-12\right)\left(\sqrt{x}+1\right)=0\)
\(\Leftrightarrow\) \(\left[{}\begin{matrix}5\sqrt{x}-12=0\\\sqrt{x}+1=0\end{matrix}\right.\) \(\Leftrightarrow\) \(\left[{}\begin{matrix}5\sqrt{x}=12\\\sqrt{x}=-1\end{matrix}\right.\) \(\Leftrightarrow\) \(\left[{}\begin{matrix}\sqrt{x}=\frac{12}{5}\\x\in\varnothing\end{matrix}\right.\)
Vậy x = \(\left\{\frac{12}{5}\right\}\)
Mình sửa chỗ cuối :
\(\sqrt{x}=\frac{12}{5}\Leftrightarrow x=\frac{144}{25}\)
