a) \(A=x^2+x+2018\)
\(A=x^2+2x.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}+2018\)
\(A=\left(x+\dfrac{1}{2}\right)^2+\dfrac{8071}{4}\)
Vì \(\left(x+\dfrac{1}{2}\right)^2\ge0\) với mọi x
\(\Rightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{8071}{4}\ge\dfrac{8071}{4}\)
=> Amin = 8071/4 <=> x + 1/2 = 0
=> x = -1/2
Vậy Amin = 8071/4 <=> x = -1/2
b) \(B=2x^2+2x+2019\)
\(B=2\left(x^2+2x.\dfrac{1}{2}+\dfrac{2019}{2}\right)\)
\(B=2\left(x^2+2x.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}+\dfrac{2019}{2}\right)\)
\(B=2\left(x^2+2x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{4037}{4}\right)\)
\(B=2\left(x^2+2x.\dfrac{1}{2}+\dfrac{1}{4}\right)+\dfrac{4037}{2}\)
\(B=2\left(x+\dfrac{1}{2}\right)^2+\dfrac{4037}{2}\)
Vì \(2\left(x+\dfrac{1}{2}\right)^2\ge0\) với mọi x
\(\Rightarrow2\left(x+\dfrac{1}{2}\right)^2+\dfrac{4037}{2}\ge\dfrac{4037}{2}\)
=> Bmin = 4037/2 <=> x + 1/2 = 0
=> x = -1/2
Vậy Bmin = 4037/2 <=> x = -1/2
c) \(C=x^2-4x+20\)
\(C=x^2-2.x.2+2^2+16\)
\(C=\left(x-2\right)^2+16\)
Vì \(\left(x-2\right)^2\ge0\) với mọi x
\(\Rightarrow\left(x-2\right)^2+16\ge16\)
=> Cmin = 16 <=> x - 2 = 0
=> x = 2
Vậy Cmin = 16 <=> x = 2
d) Bài d mình chưa nghĩ ra, sorry vì kiến thức mình không rộng 
