\(2x^2+y^2+2xy-6x-2y+10\)

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

\(=\left(x-2\right)^2+\left(x+y-1\right)^2+5\ge5\)

a) \(A=x^2+6x+10\)

\(A=x^2+2\cdot x\cdot3+3^2+1\)

\(A=\left(x+3\right)^2+1\ge1\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x+3=0\Leftrightarrow x=-3\)

b) \(B=2x^2+y^2+2xy+4x+15\)

\(B=\left(x^2+2xy+y^2\right)+\left(x^2+2\cdot x\cdot2+2^2\right)+11\)

\(B=\left(x+y\right)^2+\left(x+2\right)^2+11\ge11\forall x;y\)

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

GTNN là 2015 nha  bạn

\(B=2x^2+y^2+2xy+6x+2y+2015\)

\(=x^2+y^2+1+2xy+2y+2x+x^2+4x+4+2011\)

\(=\left(x^2+y^2+1+2xy+2y+2x\right)+\left(x^2+4x+4\right)+2011\)

\(=\left(x+y+1\right)^2+\left(x+2\right)^2+2011\)

Vì \(\left(x+y+1\right)^2+\left(x+2\right)^2\ge0\)nên \(\left(x+y+1\right)^2+\left(x+2\right)^2+2011\ge2011\)

Vậy \(MinB=2011\Leftrightarrow\hept{\begin{cases}\left(x+y+1\right)^2=0\\\left(x+2\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x+y+1=0\\x+2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-2\\y=1\end{cases}}\)

\(P=2x^2+y^2+2xy-6x-2y+10\)

\(P=\left(x^2+y^2+1^2-2y-2x\right)+\left(x^2-4x+4\right)+5\)

\(P=\left(x+y-1\right)^2+\left(x-2\right)^2+5\)

\(\left\{{}\begin{matrix}\left(x+y-1\right)^2\ge0\\\left(x-2\right)^2\ge0\end{matrix}\right.\) \(\Rightarrow P\ge5\) đẳng thức khi \(\left\{{}\begin{matrix}x-2=0\\x+y-1=0\end{matrix}\right.\) => x=2 và y=-1

2x² + y² + 2xy - 6x - 2y + 10

= x² + y² + 1² + 2xy - 2x - 2y + x² - 4x + 4 + 5

= (x + y - 1)² + (x - 2)² + 5 \(\ge\) 5

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

Vậy Min = 5 khi x = 2 và y = - 1

\(A=\left(x^2+2\cdot\dfrac{3}{2}x+\dfrac{9}{4}\right)-\dfrac{5}{4}=\left(x+\dfrac{3}{2}\right)^2-\dfrac{5}{4}\ge-\dfrac{5}{4}\\ A_{min}=-\dfrac{5}{4}\Leftrightarrow x=-\dfrac{3}{2}\\ B=\left(x^2+2xy+y^2\right)+\left(x^2+6x+9\right)+3\\ B=\left(x+y\right)^2+\left(x+3\right)^2+3\ge3\\ B_{min}=3\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=-3\end{matrix}\right.\\ C=-\left(x^2-2x+1\right)+1=-\left(x-1\right)^2+1\le1\\ C_{max}=1\Leftrightarrow x=1\)

 P=2x²+y²-2xy-6x+2y+2024

=>2P=4x²+2y²-4xy-12x+4y+4048

=(2x-y-3)²+y²-2y+1+4038

=(2x-y-3)²+(y-1)²+4038> hoặc = 4038

Dấu = xảy ra <=>2x-y-3=0 và y-1=0=>x=2;y=1=>2p=4038=>p=2019

Vậy Pmin=2019<=>x=2;y=1

Ta có: 

P = 2x² + y² - 2xy - 6x + 2y + 2024

P = (x² - 2xy + y²) - 2(x - y) + 1 + (x² - 4x + 4) + 2019

P = [(x - y)² - 2(x - y) + 1] + (x - 2)² + 2019

P = (x - y - 1)² + (x - 2)² + 2019 \(\ge\)2019 \(\forall\)x;y

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

Vậy MinP = 2019 <=> x = 2 và y = 1

\(A=2\left(x^2-2\cdot\dfrac{3}{2}x+\dfrac{9}{4}\right)-\dfrac{9}{2}=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\\ A_{min}=-\dfrac{9}{2}\Leftrightarrow x=\dfrac{3}{2}\)

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