a) \(M=x^2-8x+2018=x^2-8x+16+2002=\left(x-4\right)^2+2002\)
\(\left(x-4\right)^2\ge0\forall x\Rightarrow\left(x-4\right)^2+2002\ge2002\)
Dấu " = " xảy ra <=> x - 4 = 0 => x = 4
Vậy MMin = 2002 khi x = 4
b) \(N=4x^2-12x+2019=4x^2-12x+9+2010=\left(2x-3\right)^2+2010\)
\(\left(2x-3\right)^2\ge0\forall x\Rightarrow\left(2x-3\right)^2+2010\ge2010\)
Dấu " = " xảy ra <=> 2x - 3 = 0 => x = 3/2
Vậy NMin = 2010 khi x = 3/2
c) \(P=x^2-x+2016=x^2-x+\frac{1}{4}+\frac{8063}{4}=\left(x-\frac{1}{2}\right)^2+\frac{8063}{4}\)
\(\left(x-\frac{1}{2}\right)^2\ge0\forall x\Rightarrow\left(x-\frac{1}{2}\right)^2+\frac{8063}{4}\ge\frac{8063}{4}\)
Dấu " = " xảy ra <=> x - 1/2 = 0 => x = 1/2
Vậy PMin = 8063/4 khi x = 1/2
d) \(Q=x^2-2x+y^2+4y+2020\)
\(Q=\left(x^2-2x+1\right)+\left(y^2+4y+4\right)+2015\)
\(Q=\left(x-1\right)^2+\left(y+2\right)^2+2015\)
\(\hept{\begin{cases}\left(x-1\right)^2\ge0\forall x\\\left(y+2\right)^2\ge0\forall y\end{cases}\Rightarrow}\left(x-1\right)^2+\left(y+2\right)^2\ge0\forall x,y\)
\(\Rightarrow\left(x-1\right)^2+\left(y+2\right)^2+2015\ge2015\)
Dấu " = " xảy ra <=> \(\hept{\begin{cases}x-1=0\\y+2=0\end{cases}\Rightarrow}\hept{\begin{cases}x=1\\y=-2\end{cases}}\)
Vậy QMin = 2015 khi x = 1 ; y = -2