\(ĐK:x^2-3x+5\ge0\)
Đặt \(\sqrt{x^2-3x+5}=a\ge0\)
\(PT\Leftrightarrow a+a^2-5=7\\ \Leftrightarrow a^2+a-12=0\\ \Leftrightarrow\left(a-3\right)\left(a+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a=3\left(tm\right)\\a=-4\left(ktm\right)\end{matrix}\right.\\ \Leftrightarrow\sqrt{x^2-3x+5}=3\\ \Leftrightarrow x^2-3x+5=9\\ \Leftrightarrow x^2-3x-4=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-1\end{matrix}\right.\)
đặt \(x^2-3x=y\)
\(pt\Leftrightarrow\sqrt{y+5}+y=7\\ \Leftrightarrow\sqrt{y+5}=7-y\\ \Leftrightarrow\left\{{}\begin{matrix}y+5=\left(7-y\right)^2\\7-y\ge0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y+5=49-14y+y^2\\y\le7\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y^2-15y+44=0\\y\le7\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(y^2-11y\right)-\left(4y-44\right)=0\\y\le7\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\left(y-11\right)\left(y-4\right)=0\\y\le7\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}y=4\\y=11\end{matrix}\right.\\y\le7\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=4\\y\le7\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x^2-3x=4\\y\le7\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x^2-3x-4=0\\y\le7\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\left(x-4\right)\left(x+1\right)\\y\le7\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=4\\x=-1\end{matrix}\right.\\y\le7\end{matrix}\right.\)
Vậy \(x\in\left\{4;-1\right\}\)
