P=\(X^2+2Y^2-2XY+8X+8Y+2017\)

P=\(\dfrac{4X^2+8Y^2-8XY+32Y+32X+8068}{4}\)

P=\(\dfrac{(\sqrt{3}X)^2-2.\sqrt{3}X.\dfrac{4}{\sqrt{3}}Y+\left(\dfrac{4}{\sqrt{3}}Y\right)^2-\left(\dfrac{4}{\sqrt{3}}Y\right)^2+8Y^2+X^2+32X+32Y+8068}{4}\)

P=\(\dfrac{\left(\sqrt{3}X-\dfrac{4}{\sqrt{3}}Y\right)^2+X^2+\dfrac{8}{3}Y^2+32X+32Y+8068}{4}\)

P=\(\dfrac{\left(\sqrt{3}X-\dfrac{4}{\sqrt{3}}Y\right)^2+X^2+2.X.16+16^2+(\dfrac{2\sqrt{2}}{\sqrt{3}}Y)^2+2.\dfrac{2\sqrt{2}}{\sqrt{3}}Y.4\sqrt{6}+\left(4\sqrt{6}\right)^2+7716}{4}\)

P=\(\dfrac{\left(\sqrt{3}X-\dfrac{4}{\sqrt{3}}Y\right)^2+\left(X+16\right)^2+\left(\dfrac{2\sqrt{2}}{\sqrt{3}}Y+4\sqrt{6}\right)^2}{4}+1929\ge1929\forall X\in R\)

DẤU = XẢY RA \(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{3}X-\dfrac{4}{\sqrt{3}}Y=0\\X+16=0\\\dfrac{2\sqrt{2}}{\sqrt{3}}Y+4\sqrt{6}=0\end{matrix}\right.\)

\(x^2+2y^2+2xy-8x+32=0\)

\(\left(x^2+2xy+y^2\right)+y^2-8x+32=0\)

\(\left(x+y\right)^2-8x-8y+y^2+8y+32=0\)

\(\left(x+y\right)^2-2.\left(x+y\right).4+16+\left(y^2+2.y.4+16\right)=0\)

\(\left(x+y-4\right)^2+\left(y+4\right)^2=0\)

Vì \(\left(x+y-4\right)^2\ge0\forall x;y\)

\(\left(y+4\right)^2\ge0\forall y\)

\(\Rightarrow VT\ge0\forall x;y\)

mà \(TĐB:VT=0\)

\(\Rightarrow\hept{\begin{cases}^{ }x=8\\y=-4\end{cases}}\)

Sau đó bạn tự tính nốt nha!

Chúc bạn học tốt.

Bạn xem lại đề

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 \(\hept{\begin{cases}x=-2\\y=3\end{cases}}\)

Có link câu này bạn tham khảo xem có được không nhé

https://h.vn/hoi-dap/question/535151.html

Học tốt nhé!

Mai cho bn đấy tui dg định off =))

a)\(11x+11y-x^2-xy\)

\(=\left(11x+11y\right)-\left(x^2+xy\right)\)

\(=11\left(x+y\right)-x\left(x+y\right)\)

\(=\left(11-x\right)\left(x+y\right)\)

b)\(x^2-xy-8x+8y\)

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

\(=x\left(x-y\right)-8\left(x-y\right)\)

\(=\left(x-8\right)\left(x-y\right)\)

c)\(x^2-6x-y^2+9\)

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

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

d)\(x^2+2xy+y^2-xz-yz\)

\(=\left(x^2+2xy+y^2\right)-\left(xz+yz\right)\)

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

\(=\left(x+y\right)\left(x+y-z\right)\)

a) \(11x+11y-x^2-xy\)

\(=11\left(x+y\right)-x\left(x+y\right)\)

\(=\left(x+y\right)\left(11-x\right)\)

b) \(x^2-xy-8x+8y\)

\(=x\left(x-y\right)-8\left(x-y\right)\)

\(=\left(x-y\right)\left(x-8\right)\)

c) \(x^2-6x-y^2+9\)

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

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

\(=\left(x-3-y\right)\left(x-3+y\right)\)

d) \(x^2+2xy+y^2-xz-yz\)

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

\(=\left(x+y\right)\left(x+y-z\right)\)