\(x^2-3x-\sqrt{x^2-3x+4}+2=0\) ĐK : \(x^2-3x+4\ge0\)
\(\Leftrightarrow x^2-3x+2=\sqrt{x^2-3x+4}\)
\(\Leftrightarrow x^2-3x+4-2=\sqrt{x^2-3x+4}\)
Đặt : \(\sqrt{x^2-3x+4}=t\) \(\left(t\ge0\right)\)
\(pt\Leftrightarrow t^2-2=t\)
\(\Leftrightarrow t^2-t-2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=2\left(tm\right)\\t=-1\left(l\right)\end{matrix}\right.\)
Với \(t=2\Rightarrow\sqrt{x^2-3x+4}=2\)
\(\Leftrightarrow x^2-3x+4=4\)
\(\Leftrightarrow x^2-3x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
Ta có: \(x^2-3x-\sqrt{x^2-3x+4}+2=0\)
\(x^2-3x+4-\sqrt{x^2-3x+4}-2=0\)
Đặt \(t=\sqrt{x^2-3x+4}\left(t\ge0\right)\)
Ta có: \(t^2-t-2=0\)
\(1+\left(-2\right)-\left(-1\right)=0\)
\(\Rightarrow\)pt có 2 nghiệm.
\(\left[{}\begin{matrix}t_1=-1\left(loại\right)\\t_2=2\left(nhận\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x^2-3x+4}=2\)
\(\Leftrightarrow x^2-3x+4=4\)
\(\Leftrightarrow x^2-3x=0\)
\(\Leftrightarrow x\left(x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
Vậy nghiệm của pt là \(\left\{0;3\right\}\)
