a)9x2+12+7
Sai đề trầm trọng
b)x2-26x+180
Ta có:x2-26x+180=x2+2.13x+132+11
=(x+13)2+11
Vì (x+13)2\(\ge\)0
Suy ra:(x+13)2+11\(\ge\)11
Dấu = xảy ra khi x+13=0
x=-13
Vậy Min B=11 khi x=-13
a) \(9x^2+12x+7=\left(9x^2+12x+4\right)+3=\left(3x+2\right)^2+3\ge3\)
Min = 3 <=> x = -2/3
b) \(x^2-26x+180=\left(x^2-26x+169\right)+11=\left(x-13\right)^2+11\ge11\)
Min = 11 <=> x = 13