đk x ≥ 3
\(c,\left(\sqrt{x^2-6x+9}\right)^2=\left(4-x\right)^2\\ x^2-6x+9=16-8x+x^2\\ x^2-6x+9-16+8x-x^2=0\\ 2x-7=0\\ 2x=7\\ x=\dfrac{7}{2}\left(thoanman\right)\)
\(\sqrt[]{x^2-6x+9}=\sqrt[]{\left(x-3\right)^2}=x-3=4-x\)
\(=>2x=7=>x=\dfrac{7}{2}\)
`sqrt{x^2 -6x +9 } =4-x`
`<=> sqrt{ x^2 - 2.x.3 +3^2 } =4-x`
`<=> sqrt{(x-3)^2 } = 4-x`
`<=> |x-3| = 4-x`
`<=>`\(\left[{}\begin{matrix}x-3=4-x\\x-3=x-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+x=4+3\\x-x=-4+3\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}2x=7\\0=-1\left(vô-lí\right)\end{matrix}\right.\)
`=> 2x=7`
`=> x=7/2`
Vậy `S={7/2}`
\(\sqrt{x^2-6x+9}=4-x\)
\(\Rightarrow\left(\sqrt{x^2-6x+9}\right)^2=\left(4-x\right)^2\)
\(\Leftrightarrow x^2-6x+9=16-8x+x^2\)
\(\Leftrightarrow x^2-x^2-6x+8x=16-9\)
\(\Leftrightarrow2x=7\Leftrightarrow x=\dfrac{7}{2}\)
Vậy \(S=\left\{\dfrac{7}{2}\right\}\)
