P=2x2+y2-2xy-6x+2y+2024
=>2P=4x2+2y2-4xy-12x+4y+4048
=(2x-y-3)2+y2-2y+1+4038
=(2x-y-3)2+(y-1)2+4038> hoặc = 4038
Dấu = xảy ra <=>2x-y-3=0 và y-1=0=>x=2;y=1=>2p=4038=>p=2019
Vậy Pmin=2019<=>x=2;y=1
Ta có:
P = 2x2 + y2 - 2xy - 6x + 2y + 2024
P = (x2 - 2xy + y2) - 2(x - y) + 1 + (x2 - 4x + 4) + 2019
P = [(x - y)2 - 2(x - y) + 1] + (x - 2)2 + 2019
P = (x - y - 1)2 + (x - 2)2 + 2019 \(\ge\)2019 \(\forall\)x;y
Dấu "=" xảy ra <=> \(\hept{\begin{cases}x-y-1=0\\x-2=0\end{cases}}\) <=> \(\hept{\begin{cases}y=x-1\\x=2\end{cases}}\) <=> \(\hept{\begin{cases}y=1\\x=2\end{cases}}\)
Vậy MinP = 2019 <=> x = 2 và y = 1
\(P=2x^2+y^2-2xy-6x+2y+2024\)
\(\Rightarrow P=\left(x^2-2xy+y^2\right)-2\left(x-y\right)+1+\left(x^2-4x+4\right)+2019\)
\(\Rightarrow P=\left[\left(x-y\right)^2-2\left(x-y\right)+1\right]+\left(x-2\right)^2+2019\)
\(\Rightarrow P=\left(x-y-1\right)^2+\left(x-2\right)^2+2019\)
Ta có:
\(\left(x-y-1\right)^2\ge0\forall x;y\inℝ\)
\(\left(x-2\right)^2\ge0\forall x\inℝ\)
\(\Rightarrow\left(x-y-1\right)^2+\left(x-2\right)^2\ge0\forall x;y\inℝ\)
\(\Rightarrow\left(x-y-1\right)^2+\left(x-2\right)^2+2019\ge2019\forall x;y\inℝ\)
\(\Rightarrow P\ge2019\forall x;y\inℝ\)
Dấu "=" xảy ra:
\(\Leftrightarrow\hept{\begin{cases}x-y-1=0\\x-2=0\end{cases}\Leftrightarrow\hept{\begin{cases}y=x-1\\x=2\end{cases}\Leftrightarrow}\hept{\begin{cases}y=1\\x=2\end{cases}}}\)
Vậy P nhỏ nhất khi P = 2019 tại x=2;y=1
Chúc bạn học tốt nhé!
