giúp mình vs
P = x^2 + 2y^2 - 2xy + 8x + 8y + 2017
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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.
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) \(x^2+2xy+y^2-xz-yz\)
\(=\left(x+y\right)^2-z\left(x+y\right)\)
\(=\left(x+y\right)\left(x+y-z\right)\)
P = x2 + 2y2 - 2xy + 8x + 8y + 2017
= x2 + y2 + 42 - 2xy - 8y + 8x + y2 + 16y + 64 + 1937
= (x - y + 4)2 + (y + 8)2 + 1937 \(\ge\) 1937
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x-y+4=0\\y+8=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=-12\\y=-8\end{matrix}\right.\)
\(P=x^2+2y^2-2xy+8x+8y+2017\)
\(=\left(x^2-2xy+8x\right)+2y^2+8y+2017\)
\(=\left[x^2-2x\left(y-8\right)+\left(y-8\right)^2\right]+2y^2+8y+2017-y^2+16y-64\)\(=\left(x-y+8\right)^2+y^2+24y+1953\)
\(=\left(x-y+8\right)^2+\left(y^2+24y+144\right)+1809\)
\(=\left(x-y+8\right)^2+\left(y+12\right)^2+1809\ge1809\forall x\)Vậy Min P = 1809 khi \(\left\{{}\begin{matrix}x-y+8=0\\y+12=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x+20=0\\y=-12\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-20\\y=-12\end{matrix}\right.\)