a: \(=3\left(x^2+2x+\dfrac{5}{3}\right)\)
\(=3\left(x^2+2x+1+\dfrac{2}{3}\right)\)
\(=3\left(x+1\right)^2+2>=2\)
Dấu '=' xảy ra khi x=-1
b: Lấy x1<x2<-1
\(A=\dfrac{f\left(x_1\right)-f\left(x_2\right)}{x_1-x_2}=\dfrac{3x_1^2+6x_1-3x_2^2-6x_2}{x_1-x_2}\)
\(=3\left(x_1+x_2\right)+6\)
Vì x1<-1, x2<-1 thì x1+x2<-2
=>3(x1+x2)+6<0
=>Hàm số nghịch biến khi x<-1