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Bạn nên sửa lại đề là tìm GTNN
\(A=\left(x^2-2xy+y^2\right)+2\left(x-y\right)+1+y^2+4y+4+15\\ A=\left(x-y+1\right)^2+\left(y+2\right)^2+15\ge15\\ A_{min}=15\Leftrightarrow\left\{{}\begin{matrix}x=y-1\\y+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=-2\end{matrix}\right.\)
Vậy GTNN của A là 15
\(A=-\left(4x^2-4xy+y^2\right)-\left(y^2-2y+1\right)+4\)
\(A=-\left(2x-y\right)^2-\left(y-1\right)^2+4\)
Do \(\left\{{}\begin{matrix}-\left(2x-y\right)^2\le0\\-\left(y-1\right)^2\le0\end{matrix}\right.\) ;\(\forall x;y\)
\(\Rightarrow A\le4;\forall x;y\)
Vậy \(A_{max}=4\) khi \(x=\dfrac{1}{2};y=1\)
\(A=2\left(x^2-2xy+y^2\right)+\left(x^2-3x+\dfrac{9}{4}\right)+\dfrac{8067}{4}\)
\(A=2\left(x-y\right)^2+\left(x-\dfrac{3}{4}\right)^2+\dfrac{8067}{4}\ge\dfrac{8067}{4}\)
\(A_{min}=\dfrac{8067}{4}\) khi \(x=y=\dfrac{3}{2}\)
a)
\(A=4x-x^2+3=-\left(x^2-4x-3\right)=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\)
Daaus = xayr ra khi: x = 2
b) \(B=4x^2-12x+15=4\left(x^2-3x+9\right)-21=4\left(x-3\right)^2-21\ge-21\)
Dấu = xảy ra khi x = 3
c) \(C=4x^2+2y^2-4xy-4y+1=\left(4x^2-4xy+y^2\right)+\left(y^2-4y+4\right)-3=\left(2x-y\right)^2+\left(y-2\right)^2-3\ge-3\)
Dấu = xảy ra khi
2x = y và y = 2
=> x = 1 và y = 2
a) A = \(-x^2+4x+3=-\left(x-2\right)^2+7\le7\)
Dấu "=" <=> x = 2
b) \(4x^2-12x+15=\left(2x-3\right)^2+6\ge6\)
Dấu "=" xảy ra <=> \(x=\dfrac{3}{2}\)
c) \(4x^2+2y^2-4xy-4y+1\)
= \(\left(4x^2-4xy+y^2\right)+\left(y^2-4y+4\right)-3\)
= \(\left(2x-y\right)^2+\left(y-2\right)^2-3\ge-3\)
Dấu "=" <=> \(\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)
\(A=2\left(x^2-2xy+y^2\right)+\left(x^2-3x+\dfrac{9}{4}\right)+\dfrac{8067}{4}\)
\(A=2\left(x-y\right)^2+\left(x-\dfrac{3}{2}\right)^2+\dfrac{8067}{4}\ge\dfrac{8067}{4}\)
\(A_{min}=\dfrac{8067}{4}\) khi \(x=y=\dfrac{3}{2}\)
a) \(A=-x^2+2x=-\left(x^2-2x+1\right)+1=-\left(x-1\right)^2+1\le1\)
\(maxA=1\Leftrightarrow x=1\)
b) \(B=\left(2-3x\right)\left(3+2x\right)=-6x^2-5x+6=-6\left(x^2+\dfrac{5}{6}x+\dfrac{25}{144}\right)+\dfrac{169}{24}=-6\left(x+\dfrac{5}{12}\right)^2+\dfrac{169}{24}\le\dfrac{169}{24}\)
\(minB=\dfrac{169}{24}\Leftrightarrow x=-\dfrac{5}{12}\)
c) \(C=4xy-4x-2y-4x^2-2y^2-3=-\left[4x^2-4x\left(y-1\right)+\left(y-1\right)^2\right]+\left(y^2-4y+4\right)-6=\left(2x-y+1\right)^2+\left(y-2\right)^2-6\le-6\)
\(minC=-6\Leftrightarrow\)\(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=2\end{matrix}\right.\)
A = x^2 + 5y^2 + 4xy - 2y - 3
= x^2 + 4xy + 4y^2 + y^2 - 2y + 1 - 4
= ( x + 2y )^2 + ( y - 1 )^2 - 4 >= -4
Dấu ''='' xảy ra khi y = 1 ; x = -2
Vậy GTNN A là -4 khi x = -2 ; y = 1
\(A=-\left(4x^2-4xy+y^2\right)-\left(y^2-2y+1\right)+4\)
\(A=4-\left(2x-y\right)^2-\left(y-1\right)^2\le4\)
\(A_{max}=4\) khi \(\left\{{}\begin{matrix}x=\frac{1}{2}\\y=1\end{matrix}\right.\)