\(x^2+2.3x+9+18\\ =\left(x+3\right)^2+18\ge18\forall x\\ \)Dấu = xảy ra khi\(x+3=0\\ x=-3\)Vậy \(Min=18khix=-3\)Câu trả lời: \(x^2+2.3x+9+18\\ =\left(x+3\right)^2+18\ge18\forall x\\ \)Dấu = xảy ra khi\(x+3=0\\ x=-3\)Vậy \(Min=18khix=-3\)