\(A\left(x\right)=\left(x^2-2xy+y^2\right)+\left(x^2-2x+1\right)+2\)
\(=\left(x-y\right)^2+\left(x-1\right)^2+2\)
Do \(\left\{{}\begin{matrix}\left(x-y\right)^2\ge0\\\left(x-1\right)^2\ge0\end{matrix}\right.\) \(\forall x;y\)
\(\Rightarrow A\ge2\)
\(A_{min}=2\) khi \(\left\{{}\begin{matrix}x-y=0\\x-1=0\end{matrix}\right.\) \(\Rightarrow x=y=1\)
Các câu sau sẽ làm tắt bước trình bày, em tự trình bày đầy đủ
b.
\(B\left(x\right)=\dfrac{1}{2}\left(x^2+2xy+y^2\right)+\dfrac{1}{2}\left(x^2-2x+1\right)+\dfrac{1}{2}\left(y^2-2y+1\right)-2\left(x+y\right)-1\)
\(=\dfrac{1}{2}\left(x+y\right)^2-2\left(x+y\right)+2+\dfrac{1}{2}\left(x-1\right)^2+\dfrac{1}{2}\left(y-1\right)^2-3\)
\(=\dfrac{1}{2}\left[\left(x+y\right)-2\right]^2+\dfrac{1}{2}\left(x-1\right)^2+\dfrac{1}{2}\left(y-1\right)^2-3\ge-3\)
\(B_{min}=-3\) khi \(\left\{{}\begin{matrix}\left(x+y\right)-2=0\\x-1=0\\y-1=0\end{matrix}\right.\) \(\Rightarrow x=y=1\)
c.
\(C\left(x\right)=2\left(x^2+2xy+y^2\right)-8\left(x+y\right)+y^2+6y+9+9\)
\(=2\left(x+y\right)^2-8\left(x+y\right)+8+\left(y+3\right)^2+1\)
\(=4\left(x+y-2\right)^2+\left(y+3\right)^2+1\ge1\)
\(C_{min}=1\) khi \(\left\{{}\begin{matrix}x+y-2=0\\y+3=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=5\\y=-3\end{matrix}\right.\)
d.
\(D\left(x\right)=2\left(x^2-x+\dfrac{1}{4}\right)+3\left(y^2-\dfrac{2}{3}y+\dfrac{1}{9}\right)+4\left(z^2-\dfrac{1}{2}z+\dfrac{1}{16}\right)-\dfrac{1}{12}\)
\(=2\left(x-\dfrac{1}{2}\right)+3\left(y-\dfrac{1}{3}\right)^2+4\left(z-\dfrac{1}{4}\right)^2-\dfrac{1}{12}\ge-\dfrac{1}{12}\)
\(D_{min}=-\dfrac{1}{12}\) khi \(\left(x;y;z\right)=\left(\dfrac{1}{2};\dfrac{1}{3};\dfrac{1}{4}\right)\)
e.
\(E\left(x\right)=2x^2+8xy+11y^2-4x-2y+6\)
\(=2\left(x^2+4xy+4y^2\right)-4\left(x+2y\right)+\left(3y^2+6y+3\right)+3\)
\(=2\left(x+2y\right)^2-4\left(x+2y\right)+2+3\left(y+1\right)^2+1\)
\(=2\left(x+2y+1\right)^2+3\left(y+1\right)^2+1\ge1\)
\(E_{min}=1\) khi \(\left\{{}\begin{matrix}x+2y+1=0\\y+1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
\(2.F\left(x\right)=4x^2+12y^2+10z^2-12xy+16yz-4zx+4y+8z+4\)
\(=\left(4x^2+9y^2+z^2-12xy-4xz+6yz\right)+\dfrac{1}{3}\left[9y^2+27z^2+30yz+4\left(3y+5z\right)+4+2z^2+4z+2\right]+2\)
\(=\left(2x-3y-z\right)^2+\dfrac{1}{3}\left[\left(3y+5z\right)^2+4\left(3y+5z\right)+4+2\left(z+1\right)^2\right]+2\)
\(=\left(2x-3y-z\right)^2+\dfrac{1}{3}\left(3y+5z+2\right)^2+\dfrac{2}{3}\left(z+1\right)^2+2\ge2\)
\(\Rightarrow F\left(x\right)\ge1\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}2x-3y-z=0\\3y+5z+2=0\\z+1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=y=1\\z=-1\\\end{matrix}\right.\)
g.
\(G\left(x\right)=2x^2+2y^2+z^2+2xy-2xz-2yz-2x-4y\)
\(=\left(x^2+2xy+y^2\right)-2z\left(x+y\right)+z^2+\left(x^2-2x+1\right)+\left(y^2-4y+4\right)-5\)
\(=\left(x+y\right)^2-2z\left(x+y\right)+z^2+\left(x-1\right)^2+\left(y-2\right)^2-5\)
\(=\left(x+y-z\right)^2+\left(x-1\right)^2+\left(y-2\right)^2-5\ge-5\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x+y-z=0\\x-1=0\\y-2=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=1\\y=2\\z=3\end{matrix}\right.\)
h.
\(H\left(x\right)=x^2+y^2-xy-x+y+1\)
\(=\left(x^2-xy+\dfrac{y^2}{4}\right)-\left(x-\dfrac{y}{2}\right)+\dfrac{1}{4}+\dfrac{3y^2}{4}+\dfrac{y}{2}+\dfrac{3}{4}\)
\(=\left(x-\dfrac{y}{2}\right)^2-\left(x-\dfrac{y}{2}\right)+\dfrac{1}{4}+\dfrac{3}{4}\left(y^2+\dfrac{2}{3}y+\dfrac{1}{9}\right)+\dfrac{2}{3}\)
\(=\left(x-\dfrac{y}{2}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\left(y+\dfrac{1}{3}\right)^2+\dfrac{2}{3}\ge\dfrac{2}{3}\)
\(H_{min}=\dfrac{2}{3}\) khi \(\left\{{}\begin{matrix}x-\dfrac{y}{2}-\dfrac{1}{2}=0\\y+\dfrac{1}{3}=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}\\y=-\dfrac{1}{3}\end{matrix}\right.\)
