Question

Tìm x,y để các biểu thức sau đạt GTNN, tính giá trị đó:

a) \(M=x^2-8x+2018\)

b) \(N=4x^2-12x+2019\)

c) \(P=x^2-x+2016\)

d) \(Q=x^2-2x+y^2+4y+2020\)

Answer 1

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 M_Min = 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 N_Min = 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 P_Min = 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 Q_Min = 2015 khi x = 1 ; y = -2