`x^2 + 6x + 6 = 0`
`=> x(x + 6) = -6`
`=> -6 vdots x`
`=> x∈Ư(-6) = {1;-1;2;-2;3;-3;6;-6}`
`=> x - 6 ∈ {-6;6;-3;3;-2;2;-1;1}`
`=> x ∈ {7;5;8;4;9;3;12;0}`
=> Ko tồn tại x
x2 + 6x + 6
\(\Leftrightarrow\left(x^2+6x+9\right)-3=0\)
\(\Leftrightarrow\left(x+3\right)^2-3=0\)
\(\Leftrightarrow\left(x+3-\sqrt{3}\right)\left(x+3+\sqrt{3}\right)=0\)
\(\Rightarrow\) +) TH1 : x = \(\left(\sqrt{3}\right)-3\)
+) TH2 : x = \(\left(-\sqrt{3}\right)-3\)
Ta có: \(x^2+6x+6=0\)
\(\Leftrightarrow x^2+6x+9-3=0\)
\(\Leftrightarrow\left(x+3\right)^2=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=\sqrt{3}\\x+3=-\sqrt{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{3}-3\\x=-\sqrt{3}-3\end{matrix}\right.\)
Vậy: \(S=\left\{\sqrt{3}-3;-\sqrt{3}-3\right\}\)
