a/ \(2x^2+12x+21=2\left(x^2+6x+9\right)+3=2\left(x+3\right)^2+3\ge3\)
Min = 3 <=> x = -3
b/ \(9x^2-30x+26=9\left(x-\frac{5}{3}\right)^2+1\ge1\)
Min = 1 <=> x = 5/3
a)2x2+12x+21
Ta có:2x2+12x+21=2.(x2+6x+32)+3
=2.(x+3)2+3
Vì 2.(x+3)2\(\ge\)0
Suy ra:2.(x+3)2+3\(\ge\)3
Dấu = xảy ra khi x+3=0
x=-3
Vậy MinA=3 khi x=-3
b)9x2-30x+26
Ta có:9x2-30x+26=(3x)2-2.15x+52+1
=(3x-5)2+1
Vì (3x-5)2\(\ge\)0
Suy ra:(3x-5)2+1\(\ge\)1
Dấu = xảy ra khi 3x-5=0
3x=5
x=\(\frac{5}{3}\)
Vậy Min B=1 khi x=\(\frac{5}{3}\)
