A=x²+2y²+2xy+2x-4y+2013

=x²+y²+1+2xy+2x+2y+y²-6y+9+2003

=(x+y+1)²+(y-3)²+2003

Min A=2003 tại x=-4;y=3

A= (X²+2XY+Y²) + 2(X+Y)+1+Y²-6Y+9+2003

A=(X+Y)²+ 2(X+Y)+1+(Y-3)²+2003

A=(X+Y+1)²+(Y-3)²+2003

=> A>=2003

(DẤU "=" XẢY RA KHI X=-4;Y=3)

lớn nhất chứ

tích đúng mình làm cho

là sao ạk
giải giùm mình với ạk

Ta có

\(A=x^2+2y^2+2xy-2x-8y+2017\)

\(=\left(x^2+2xy+y^2\right)-2\left(x+y\right)+1+\left(y^2-6y+9\right)+2007\)

\(=\left(x+y\right)^2-2\left(x+y\right)+1+\left(y-3\right)^2+2007\)

\(=\left(x+y-1\right)^2+\left(y-3\right)^2+2007\ge2007\)

Dấu = xảy ra khi \(\hept{\begin{cases}x=-2\\y=3\end{cases}}\)

\(P=x^2+2y^2+2xy-6x-4y+13\)

\(=\left(x^2+2xy+y^2\right)+y^2-6\left(x+y\right)+2y+13\)

\(=\left(x+y\right)^2-2\left(x+y\right)3+9+y^2+2y+1+3\)

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

Mà \(\left(x+y-3\right)^2\ge0\forall x;y\)

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

\(\Rightarrow P\ge3\forall x;y\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x+y-3=0\\y+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=4\\y=-1\end{matrix}\right.\)

Vậy \(P_{Min}=3\Leftrightarrow\left(x;y\right)=\left(4;-1\right)\)

Ta có: P = x² + 2y² + 2xy - 6x -4y +13

= (x² + y² + 9 + 2xy - 6x - 6y) + (y² + 2y + 1) + 3

= (x + y - 3)² + (y + 1)² + 3

Ta thấy (x + y - 3)² ≥ 0 với mọi x,y

(y + 1)² ≥ 0 với mọi x,y

⇔ (x + y - 3)² + (y + 1)² ≥ 0 với mọi x,y

⇔ (x + y - 3)² + (y + 1)² +3 ≥ 3 với mọi x,y

hay P ≥ 3 với mọi x,y

Dấu "=" xảy ra

⇔ \(\left\{{}\begin{matrix}x+y-3=0\\y+1=0\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}x+y-3=0\\y=-1\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}x-1-3=0\\y=-1\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}x=4\\y=-1\end{matrix}\right.\)

Vậy GTNN của biểu thức P là 3 khi x=4 và y=-1.

2C=4x^2+2x-10=((2x)^2+4x\(\dfrac{1}{2}\)+\(\dfrac{1}{4}\))-\(\dfrac{41}{4}\)

=\(\left(2x+\dfrac{1}{2}\right)^2\)-41/4\(\ge\dfrac{-41}{4}\)

=> C\(\ge\dfrac{-41}{8}\)

Vậy min C = \(\dfrac{-41}{8}\)khi x=\(\dfrac{-1}{4}\)

\(4A=4x^2+44y^2+24xy-8y+20=\left(2x\right)^2+2.2x.6y+\left(6y\right)^2+8y^2-8y+20=\left(2x+6y\right)^2+2\left(4y^2-4y+1\right)+18=\left(2x+6y\right)^2+2\left(2y-1\right)^2+18\ge18\)

\(A=\left(x^2+y^2+1+2xy+2x+2y\right)+\left(y^2-6y+9\right)+2007\)

\(A=\left(x+y+1\right)^2+\left(y-3\right)^2+2007\ge2007\)

\(A_{min}=2007\) khi \(\left\{{}\begin{matrix}x=-4\\y=3\end{matrix}\right.\)

b/ Đề sai, cho \(x=y=1\Rightarrow5\ge6\)

\(A=\left(x^2+2xy+y^2\right)+\left(2x+2y\right)+1+\left(y^2-6y+9\right)+2006\)\(=\left(x+y\right)^2+2\left(x+y\right)+1+\left(y-3\right)^2+2006\)

\(=\left(x+y+1\right)^2+\left(y-3\right)^2+2006\)

Ta có: \(\left(x+y+1\right)^2+\left(y-3\right)^2\ge0\left(\forall x;y\right)\)

\(\Rightarrow A\ge2006\).

Vậy MIN A = 2006 \(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y+1\right)^2=0\\\left(y-3\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-4\\y=3\end{matrix}\right.\)