2) a) \(P=3x^2+y^2-8x+2xy+16\)

\(P=\left(x^2+2xy+y^2\right)+2\left(x^2-4x+4\right)+8\)

\(P=\left(x+y\right)^2+2\left(x-2\right)^2+8\ge8\forall x;y\)

\(\Rightarrow\) GTNN của P là 8 khi \(\left\{{}\begin{matrix}\left(x+y\right)^2=0\\\left(x-2\right)^2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=-x\\x=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-2\end{matrix}\right.\) vậy GTNN của P là 8 khi \(x=2;y=-2\)

b) \(Q=x^2+2y^2-2xy-4y+2017\)

\(Q=\left(x^2-2xy+y^2\right)+\left(y^2-4y+4\right)+2013\)

\(Q=\left(x-y\right)^2+\left(y-2\right)^2+2013\ge2013\forall x;y\)

\(\Rightarrow\) GTNN của Q là 2013 khi \(\left\{{}\begin{matrix}\left(x-y\right)^2=0\\\left(y-2\right)^2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=y\\y=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=2\end{matrix}\right.\) vậy GTNN của Q là 2013 khi \(x=y=2\)

c) \(M=2x^2+y^2-2xy-2x+2016\)

\(M=\left(x^2-2xy+y^2\right)+\left(x^2-2x+1\right)+2015\)

\(M=\left(x-y\right)^2+\left(x-1\right)^2+2015\ge2015\forall x;y\)

\(\Rightarrow\) GTNN của M là 2015 khi \(\left\{{}\begin{matrix}\left(x-y\right)^2=0\\\left(x-1\right)^2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\x-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=y\\x=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\) vậy GTNN của M là 2015 khi \(x=y=1\)

thanks bạn

\(P=3x^2+y^2-8x+2xy+16\)

\(P=\left(x^2+2xy+y^2\right)+\left(2x^2-8x+8\right)+8\)

\(P=\left(x+y\right)^2+2\left(x-2\right)^2+8\ge8\)

Vậy GTNN của P=8 <=> \(\orbr{\begin{cases}x+y=0\\x-2=0\end{cases}}\)<=>\(\orbr{\begin{cases}y=-2\\x=2\end{cases}}\)

Bài 2:

C=A-B

\(=2x^2-6xy+4y^2+5x^2-4xy-7y^2\)

\(=7x^2-10xy-3y^2\)

\(=7\cdot1^2-10\cdot1\cdot\dfrac{1}{2}-3\cdot\dfrac{1}{4}=7-5-\dfrac{3}{4}=2-\dfrac{3}{4}=\dfrac{5}{4}\)

Ta có \(C=x^2+2y^2-2xy-4y+5=\left(x-2xy+y^2\right)+\left(y^2-4y+4\right)+1\)

\(=\left(x-y\right)^2+\left(y-2\right)^2+1\)

Do \(\left(x-y\right)^2\ge0;\left(y-2\right)^2\ge0\Rightarrow C\ge1\)

Vậy GTNN của C là 1 khi \(\hept{\begin{cases}x-y=0\\y-2=0\end{cases}}\Leftrightarrow x=y=2\)

\(Q=x^2+2y^2-2xy-4y+2017\)

\(Q=\left(x^2-2xy+y^2\right)+\left(y^2-4y+4\right)+2013\)

\(Q=\left(x-y\right)^2+\left(y-2\right)^2+2013\ge2013\)

Vậy GTNN của Q=2013 <=> \(\orbr{\begin{cases}x-y=0\\y-2=0\end{cases}}\)<=>\(\orbr{\begin{cases}\\\end{cases}}x=y=2\)