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-x2-4xy-5y2+6y+1672
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HK
1
MA
27 tháng 9 2016
\(A=-x^2-4xy-5y^2+6y+1672\)
\(A=-x^2-4xy-4y^2-y^2+6y-9+1681\)
\(A=-\left(x+2y\right)^2-\left(y-3\right)^2+1681\)
\(A=1681-\left[\left(x+2y\right)^2+\left(y-3\right)^2\right]\)
Có: \(\left(x+2y\right)^2+\left(y-3\right)^2\ge0\)
\(\Rightarrow1681-\left[\left(x+2y\right)^2+\left(y-3\right)^2\right]\le1681\)
Dấu = xảy ra khi: \(\left(x+2y\right)^2+\left(y-3\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}\left(x+2y\right)^2=0\\\left(y-3\right)^2=0\end{cases}}\Rightarrow\hept{\begin{cases}x+2y=0\\y-3=0\end{cases}}\Rightarrow\hept{\begin{cases}x+2y=0\\y=3\end{cases}}\Rightarrow\hept{\begin{cases}x=-6\\y=3\end{cases}}\)
Vậy: \(Max_A=1681\) tại \(\hept{\begin{cases}x=-6\\y=3\end{cases}}\)
NK
1
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 12 2020
\(C=\left(x^2+4y^2+25-4xy+10x-20y\right)+\left(y^2-2y+1\right)+2\)
\(C=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\)
\(C_{min}=2\) khi \(\left\{{}\begin{matrix}x=-3\\y=1\end{matrix}\right.\)
15 tháng 4 2017
\(A=-x^2-4xy-5y^2-6y+1672\)
\(=-x^2-4xy-4y^2-y^2-6y-9+1681\)
\(=-\left(x^2+4xy+4y^2\right)-\left(y^2+6y+9\right)+1681\)
\(=-\left(x+2y\right)^2-\left(y+3\right)^2+1681\)
Dễ thấy: \(\left\{{}\begin{matrix}\left(x+2y\right)^2\ge0\forall x,y\\\left(y+3\right)^2\ge0\forall y\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}-\left(x+2y\right)^2\le0\forall x,y\\-\left(y+3\right)^2\le0\forall y\end{matrix}\right.\)
\(\Rightarrow-\left(x+2y\right)^2-\left(y+3\right)^2\le0\forall x,y\)
\(\Rightarrow A=-\left(x+2y\right)^2-\left(y+3\right)^2+1681\le1681\forall x,y\)
Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}-\left(x+2y\right)^2=0\\-\left(y+3\right)^2=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x+2y=0\\y+3=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=-2y\\y=-3\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=6\\y=-3\end{matrix}\right.\)
K
2
22 tháng 7 2021
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
MA
2
7 tháng 11 2021
ko có 2x2 đâu mik thấy đề bài nó ghi như thế. bn giúp mik nhé!
K
3
H9
HT.Phong (9A5)
CTVHS
4 tháng 8 2023
a) \(M=x^2-3x+10\)
\(M=x^2-2\cdot\dfrac{3}{2}\cdot x+\dfrac{9}{4}+\dfrac{31}{4}\)
\(M=\left(x^2-2\cdot\dfrac{3}{2}\cdot x+\dfrac{9}{4}\right)+\dfrac{31}{4}\)
\(M=\left(x-\dfrac{3}{2}\right)^2+\dfrac{31}{4}\)
Mà: \(\left(x-\dfrac{3}{2}\right)^2\ge0\) nên: \(M=\left(x-\dfrac{3}{2}\right)^2+\dfrac{31}{4}\ge\dfrac{31}{4}\)
Dấu "=" xảy ra
\(\left(x-\dfrac{3}{2}\right)^2+\dfrac{31}{4}=\dfrac{31}{4}\Leftrightarrow\left(x-\dfrac{3}{2}\right)^2=0\)
\(\Leftrightarrow x-\dfrac{3}{2}=0\Leftrightarrow x=\dfrac{3}{2}\)
Vậy: \(M_{min}=\dfrac{31}{4}\) với \(x=\dfrac{3}{2}\)
b) \(N=2x^2+5y^2+4xy+8x-4y-100\)
\(N=x^2+x^2+4y^2+y^2+4xy+8x-4y-120+16+4\)
\(N=\left(x^2+4xy+4y^2\right)+\left(x^2+8x+16\right)+\left(y^2-4y+4\right)-120\)
\(N=\left(x+2y\right)^2+\left(x+4\right)^2+\left(y-2\right)^2-120\)
Mà:
\(\left\{{}\begin{matrix}\left(x+2y\right)^2\ge0\\\left(x+4\right)^2\ge0\\\left(y-2\right)^2\ge0\end{matrix}\right.\) nên \(N=\left(x+2y\right)^2+\left(x+4\right)^2+\left(y-2\right)^2-120\ge120\)
Dấu "=" xảy ra:
\(\left\{{}\begin{matrix}\left(x+2y\right)^2=0\\\left(x+4\right)^2=0\\\left(y-2\right)^2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}-4+2y=0\\x=-4\\y=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=-4\\y=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-4\\y=2\end{matrix}\right.\)
Vậy: \(N_{min}=120\) khi \(\left\{{}\begin{matrix}x=-4\\y=2\end{matrix}\right.\)
22 tháng 12 2021
\(A=\left(x^2-6x+9\right)+2=\left(x-3\right)^2+2\ge2\\ A_{min}=2\Leftrightarrow x=3\\ B=2\left(x^2-10x+25\right)+51=2\left(x-5\right)^2+51\ge51\\ B_{min}=51\Leftrightarrow x=5\\ C=\left[\left(x^2-4xy+4y^2\right)+10\left(x-2y\right)+25\right]+\left(y^2-2y+1\right)+2\\ C=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\\ C_{min}=2\Leftrightarrow\left\{{}\begin{matrix}x-2y+5=0\\y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2y-5=2-5=-3\\y=1\end{matrix}\right.\)
22 tháng 12 2021
a) \(A=\left(x^2-6x+9\right)+2=\left(x-3\right)^2+2\ge2\)
\(minA=2\Leftrightarrow x=3\)
b) \(B=2\left(x^2-10x+25\right)+51=2\left(x-5\right)^2+51\ge51\)
\(minB=51\Leftrightarrow x=5\)
c) \(C=\left[x^2-2x\left(2y-5\right)+\left(2y-5\right)^2\right]+\left(y^2-2y+1\right)+2=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\)
\(minC=2\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=1\end{matrix}\right.\)
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mơn bb*