Bài 1:
Dự đoán điểm rơi: \(x=1;y=3\).
\(A=\left(x-y\right)^2+\left(x+1\right)^2+\left(y-5\right)^2+2021\)
\(=x^2-2xy+y^2+x^2+2x+1+y^2-10y+25+2021\)
\(=\left(x^2-2x+1\right)+\left(y^2-6y+9\right)+\left(x^2+y^2+4+4x-4y-2xy\right)+2021+25-9\)
\(=\left(x-1\right)^2+\left(y-3\right)^2+\left(x-y-2\right)^2+2023\)\(\ge2023\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(y-3\right)^2=0\\\left(x-y-2\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\end{matrix}\right.\)
Vậy \(MinA=2023\).
\(B=\left(x+2y\right)^2+\left(x-4\right)^2+\left(y-1\right)^2-27\)
\(=x^2+4xy+4y^2+x^2-8x+16+y^2-2y+1-27\)
\(=2\left(x^2+y^2+4+2xy-4x-4y\right)+3\left(y^2+2y+1\right)-27-2.4-3.1+16+1\)
\(=2\left(x+y-2\right)^2+3\left(y+1\right)^2-21\ge-21\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y-2\right)^2=0\\\left(y+1\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=-1\end{matrix}\right.\)
Vậy \(MinB=-21\)
\(C=x^4-8xy-x^3y+x^2y^2-xy^3+y^4+2022\)
\(=x^3\left(x-y\right)-y^3\left(x-y\right)+\left(x^2y^2-8xy+16\right)+2006\)
\(=\left(x-y\right)\left(x^3-y^3\right)+\left(xy-4\right)^2+2006\)
\(=\left(x-y\right)^2\left(x^2+xy+y^2\right)+\left(xy-4\right)^2+2006\)
Ta có: \(x^2+xy+y^2>0\forall x,y\)
\(\Rightarrow A\ge2006\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)^2\left(x^2+xy+y^2\right)=0\\\left(xy-4\right)^2=0\end{matrix}\right.\Leftrightarrow x=y=2\)
Vậy \(MinC=2006\)
2
\(M=\left(x-2y+1\right)^2+\left(2x+ay+5\right)^2\ge0\)
\(dấu"="\Leftrightarrow\left\{{}\begin{matrix}x-2y+1=0\\2x+ay+5=0\end{matrix}\right.\) \(có\) \(nghiệm\)\(\Leftrightarrow2\ne\dfrac{a}{-2}\Leftrightarrow a\ne4\Rightarrow min=0\)
\(a=-4\Rightarrow M=\left(x-2y+1\right)^2+\left(2x-4y+5\right)^2\)
\(=\left(x-2y+1\right)^2+4\left(x-2y+\dfrac{5}{2}\right)^2\)
\(đặt:x-2y+1=t\Leftrightarrow M=t^2+4\left(t+\dfrac{3}{2}\right)^2=5t^2+12t+9=5\left(t+\dfrac{6}{5}\right)^2+\dfrac{9}{5}\ge\dfrac{9}{5}\)
\(dấu"="\Leftrightarrow t=-\dfrac{6}{5}=x-2y+1\)
\(\Leftrightarrow x-2y=-\dfrac{11}{5}\Rightarrow\left(x;y\right)=\left(b;\dfrac{b+\dfrac{11}{5}}{2}\right)\)\(\Rightarrow min=\dfrac{9}{5}\)
