\(A=x^2-6x+11=x^2-2x.3+9+2=\left(x-3\right)^2+2\)Ta có \(\left(x-3\right)^2\ge0\Leftrightarrow\left(x-3\right)^2+2\ge2\). Vậy min A=2 khí=3. \(D=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)=\left[\left(x+1\right)\left(x+6\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]=\left(x^2+7x+6\right)\left(x^2+5x+6\right)=x^4+5x^3+6x^2+7x^3+35x^2+42x+6x^2+30x+36=x^4+12x^3+47x^2+72x+36\)(Làm ko đúng đâu, nói chung min D=0 khi x=-1
A=(x^2-6x+9)+2
=(x+3)^2+2
do (x+3)^2>=0 .nên (x+3)^2+2>=0
dấu '=' xảy ra khi x+3=0 hay x=-3
vậy min A=2 tại x=-3
