Có A = x^2 +y^2-xy-x+y+1
=> 2A =2x^2 + 2y^2 -2xy -2x +2y+2 =(x^2 -2xy +y^2)+ (x^2 -2x+1) +(y^2 +2y +1) =(x-y)^2 +(x-1)^2 +(y+1)^2 >=0
=> Min A =0
Còn lại bạn tự giải nka!@
\(B=\left(x-2\right)^2+\left(y+1\right)^2-2017\ge-2017\forall x;y\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x-2=0\\y+1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\y=-1\end{cases}}}\)
Vậy...
\(-----Hd----\)
\(Taco:\left(x-2\right)^2,\left(y+1\right)^2\ge0\forall x,y\)
\(\Rightarrow B=-2017+\left(x-2\right)^2+\left(y+1\right)^2\ge-2017\)
\(\Rightarrow B_{min}=-2017\)
Dấu "=" xảy ra khi:
\(\hept{\begin{cases}x-2=0\Leftrightarrow x=2\\y+1=0\Leftrightarrow y=-1\end{cases}}\)
Vậy GTNN của B là: -2017 khi: x=2 và y=-1
\(B=-2017+\left(x-2\right)^2+\left(y+1\right)^2\)
\(\text{Vì }\left(x-2\right)^2,\left(y+1\right)^2\ge0\forall x\)
\(\Rightarrow B=-2017+\left(x-2\right)^2+\left(y+1\right)^2\ge-2017\)
Dấu '' = '' xảy ra khi :
\(\orbr{\begin{cases}\left(x-2\right)^2=0\\\left(y+1\right)^2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\y=-1\end{cases}}\)
Vậy MinB = -2017 <=> x = 2 , y = -1
