函数y=(x2+4x+5)(x2+4x+2)+2x2+8x+1的最小值是x后的2是平方

Question

函数y=(x2+4x+5)(x2+4x+2)+2x2+8x+1的最小值是
x后的2是平方

Answer 1

设t=x^2+4x+4=(x+2)^2
y=(t+1)(t-2)+2(t-4)+1
=(t+1)(t-2)+2t-8+1
=t^2-t-2+2t-7
=t^2+t-9
=t^2+t+1/4-37/4
=(t+1/2)^2-37/4
最小值＝（0+1/2)^2-37/4= -9