\(x^2-6x+3=\sqrt{x+3}\)
\(\Leftrightarrow\left(\sqrt{x+3}-3\right)-\left(x^2-6x\right)=0\)
\(\Leftrightarrow\dfrac{x+3-9}{\sqrt{x+3}+3}-x\left(x-6\right)=0\)
\(\Leftrightarrow\left(\dfrac{1}{\sqrt{x+3}+3}-x\right)\left(x-6\right)=0\)
Trường hợp 1:
\(x-6=0\)
\(\Leftrightarrow x=6\)
Trường hợp 2:
\(\dfrac{1}{\sqrt{x+3}+3}-x=0\)
\(\Leftrightarrow x\sqrt{x+3}=1-3x\)
\(\Leftrightarrow x^3+3x^2=1-6x+9x^2\)
\(\Leftrightarrow x^3-6x^2+6x-1=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-5x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-\dfrac{5+\sqrt{21}}{2}\right)\left(x-\dfrac{5-\sqrt{21}}{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(l\right)\\x=\dfrac{5+\sqrt{21}}{2}\left(l\right)\\x=\dfrac{5-\sqrt{21}}{2}\left(n\right)\end{matrix}\right.\)
Vậy phương trình có 2 nghiệm phân biệt . . .
