P = x^2 +2y^2 - 2xy + 8x + 8y + 2017
Q = - x^2 - 2y^2 - 2xy + 8x + 6y + 13
E = -x^2 - 4y^2 + 2xy + 2x + 10xy - 3
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LC
2
PA
28 tháng 7 2017
Q = - x2 - 2y2 - 2xy + 8x + 6y + 13
= - x2 - y2 - 42 - 2xy + 8y + 8x - y2 - 2y - 1 + 30
= 30 - (x2 + y2 + 42 + 2xy - 8y - 8x) - (y2 + 2y + 1)
= 78 - (x + y - 4)2 - (y + 1 )2 \(\le\) 30
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x+y-4=0\\y+1=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=-1\end{matrix}\right.\)
TN
28 tháng 7 2017
\(Q=-x^2-2y^2-2xy+8x+6y+13\)
\(=-\left(x^2+2xy-8x\right)-2y^2+6y+13\)
\(=-\left[x^2+2x\left(y-8\right)+\left(y-8\right)^2\right]-2y^2+6y+13-\left(y-8\right)^2\)\(=-\left(x+y-8\right)^2-2y^2+6y+13-y^2+16y-64\)\(=-\left(x+y-8\right)^2-3y^2+22y-51\)
\(=-\left(x+y-8\right)^2-3\left(y^2-\dfrac{22}{3}y+\dfrac{121}{9}\right)-\dfrac{32}{3}\)\(=-\left(x+y-8\right)^2-3\left(y-\dfrac{11}{9}\right)^2-\dfrac{32}{3}\le-\dfrac{32}{2}\forall x\)Vậy Max Q = \(-\dfrac{32}{3}\) khi \(\left\{{}\begin{matrix}x+y-8=0\\y-\dfrac{11}{9}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-\dfrac{61}{9}=0\\y=\dfrac{11}{9}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{61}{9}\\y=\dfrac{11}{9}\end{matrix}\right.\)
BM
0
20 tháng 6 2017
a ) \(x^2-x+1\)
\(\Leftrightarrow\left(x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right)+\dfrac{3}{4}\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Ta có : \(\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
Vậy GTNN là \(\dfrac{3}{4}\Leftrightarrow x=\dfrac{1}{2}.\)
B
8 tháng 7 2019
\(A=\sqrt{2x^2-4x+3}+3\)
Ta có: \(2x^2-4x+3\)
\(=2\left(x^2-2x+\frac{3}{2}\right)\)
\(=2\left(x^2-2.x.1+1^2+\frac{1}{2}\right)\)
\(=2[\left(x-1\right)^2+\frac{1}{2}]\)
\(=2\left(x-1\right)^2+1\ge1\)
\(\Rightarrow\sqrt{2\left(x-1\right)^2+1}\ge\sqrt{1}\)
\(\Rightarrow\sqrt{2\left(x-1\right)^2+1}+3\ge3+\sqrt{1}=4\)
\(\Rightarrow MinA=4\Leftrightarrow x=1\)
NP
1
NN
0
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 \(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{3}X-\dfrac{4}{\sqrt{3}}Y=0\\X+16=0\\\dfrac{2\sqrt{2}}{\sqrt{3}}Y+4\sqrt{6}=0\end{matrix}\right.\)