a)2x^2+xy-y^2-x+2y-1

=2x^2+xy-x-(y-1)^2

=2x^2+x(y-1)-(y-1)^2

=2a^2+ab-b^2         với a=x,b=y-1

=2a^2+2ab-ab-b^2

=(2a-b)(a+b)

=(2x-y+1)(x+y-1)

\(B=x^2+2y^2-2xy+2x-4y-12\)

\(B=\left(x^2-2xy+y^2\right)+y^2+2x-4y-12\)

\(B=\left[\left(x-y\right)^2+2\left(x-y\right)+1\right]+\left(y^2-2y+1\right)+10\)

\(B=\left(x-y+1\right)^2+\left(y-1\right)^2+10\)

Mà  \(\left(x-y+1\right)^2\ge0\forall x;y\)

       \(\left(y-1\right)^2\ge0\forall y\)

\(\Rightarrow B\ge10\)

Dấu "=" xảy ra khi :  \(\hept{\begin{cases}x-y+1=0\\y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0\\y=1\end{cases}}\)

Vậy  \(B_{Min}=10\Leftrightarrow\left(x;y\right)=\left(0;1\right)\)

Sai rồi bạn

\(B=x^2+2y^2-2xy+2x-10y+2028\\ =x^2+y^2+1-2xy+2x-2y+y^2-8y+16+2011\\ =\left(x-y+1\right)^2+\left(y-4\right)^2+2011\ge2011\)

vì \(\left(x-y+1\right)^2\ge0;\left(y-4\right)^2\ge0\)

min B = 2011 khi \(\left\{{}\begin{matrix}y-4=0\\x-y+1=0\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}y=4\\x=3\end{matrix}\right.\)

2A = 4x² + 4xy + 4y² - 12x - 12y + 8040

= (2x + y)² - 6(2x + y) + 9 + 3y² - 6y + 3 + 8028

= (2x + y - 3)² + 3(y - 1)² + 8028 \(\ge8028\)

=> \(A\ge4014\)

Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}2x+y=3\\y-1=0\end{matrix}\right.\Leftrightarrow x=y=1\)

Vậy Min A = 4014 khi x = y = 1

\(\sqrt{x^2+y^2-2xy+2x-2y+5}+2y^2-8y+2015\)

\(=\sqrt{\left(x^2+y^2-2xy\right)+2\left(x-y\right)+1+4}+2\left(y^2-4y+4\right)+2007\)\(=\sqrt{\left(x-y+1\right)^2+4}+2\left(y-2\right)^2+2007\ge2007\)

Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}x-y+1=0\\y-2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)

Theo như bài của bạn thì GTNN là 2009 đấy

A=(5x-3y-2)² + (x+y+1)² + 4

Vậy giá trị nhỏ nhất của A là 4

\(D=x^2+y^2+z^2-2xy+2zx-2yz+y^2+2z^2+2yz-2\left(x-y+z\right)-4y-6z+19\)

\(=\left(x-y+z\right)^2-2\left(x-y+z\right)+1+\left(y^2+z^2+2yz-4y-4z+4\right)+z^2-2z+1+13\)

\(=\left(x-y+z-1\right)^2+\left(y+z-2\right)^2+\left(z-1\right)^2+13\ge13\)

\(D_{min}=13\) khi \(\left\{{}\begin{matrix}x-y+z=1\\y+z=2\\z=1\end{matrix}\right.\) \(\Rightarrow x=y=z=1\)

\(M=x^2+2y^2+2xy-2x-3y+1\)

=> \(M=x^2+2x\left(y-1\right)+\left(y-1\right)^2-\left(y-1\right)^2+2y^2-3y+1\)

=> \(M=\left(x+y-1\right)^2-y^2+2y-1+2y^2-3y+1\)

=> \(M=\left(x+y-1\right)^2+y^2-y\)

=> \(M=\left(x+y-1\right)^2+y^2-2y\frac{1}{2}+\frac{1}{4}-\frac{1}{4}\)

=> \(M=\left(x+y-1\right)^2+\left(y-\frac{1}{2}\right)^2-\frac{1}{4}\)

Có \(\left(x+y-1\right)^2\ge0\)với mọi x, y

\(\left(y-\frac{1}{2}\right)^2\ge0\)với mọi y

=> \(M=\left(x+y-1\right)^2+\left(y-\frac{1}{2}\right)^2-\frac{1}{4}\ge\frac{-1}{4}\)với mọi x, y

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x+y-1=0\\y-\frac{1}{2}=0\end{cases}}\)

<=> \(\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{1}{2}\end{cases}}\)

KL: M_min = \(\frac{-1}{4}\)<=> \(x=y=\frac{1}{2}\)

cảm ơn Giang