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\(A=-x^2-4xy-5y^2+6y+1672\)
\(A=-x^2-4xy-4y^2-y^2+6y-9+1681\)
\(A=-\left(x+2y\right)^2-\left(y-3\right)^2+1681\)
\(A=1681-\left[\left(x+2y\right)^2+\left(y-3\right)^2\right]\)
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 \(\hept{\begin{cases}x=-6\\y=3\end{cases}}\)
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Ta có :
\(D=-x^2-4xy-5y^2+6y+1672\)
\(=-\left(x^2+4xy+4y^2\right)-\left(y^2+6y+9\right)+9+1672\)
\(=-\left(x+2y\right)^2-\left(y+3\right)^2+1681\)
Có :
\(\left(x+2y\right)^2\ge0\)
\(\left(y+3\right)^2\ge0\)
\(\Rightarrow-\left(x+2y\right)^2-\left(y+3\right)^2+1681\le1681\)
\(\Rightarrow Max_N=1681\Leftrightarrow\hept{\begin{cases}y=-3\\x=6\end{cases}}\)
Vậy ...
-(x-2y)^2 -(y-3)^2 +1681
Với mọi x, y ta có: -....<=0
=>-.... <= 1681
Dấu = xảy ra khi
x=2y; y=3
=> x=6;y=3
Vậy...
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2+5y^2-4xy-6y+9=0\)
\(\Leftrightarrow\left(x^2-4xy+4y^2\right)+\left(y^2-6y+9\right)=0\)
\(\Leftrightarrow\left(x-2y\right)^2+\left(y-3\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2y=0\\y-3=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=6\\y=3\end{matrix}\right.\)
\(\Rightarrow A=...\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) A = x2 - 6x + 13 = x2 - 2.x.3 + 33 +4 = (x-3)2 + 4 >= 4 suy ra minA=4
mấy câu kia giải tương tự
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1. x2-4xy + 5y2 = 100\(\Leftrightarrow\left(x^2-4xy+4y^2\right)+y^2=100\)
\(\Leftrightarrow\left(x-2y\right)^2+y^2=0+10^2=6^2+8^2\)\(\Leftrightarrow\int^{x-2y=0}_{y=10}\)
hoặc \(\int^{x-2y=10}_{y=0}\) hoặc \(\int^{x-2y=6}_{y=8}\) hoặc \(\int^{x-2y=8}_{y=6}\)
từ đó ta tìm được (x;y)= ( 20;10);(10;0) ; ( 24;6) ; ( 20; 6)
2. 4x2 + 2y2 - 4xy + 20x - 6y + 29 = 0 \(\Leftrightarrow4x^2-4x\left(y-5\right)+\left(y^2-10y+25\right)+\left(y^2+4y+4\right)=0\)
\(\Leftrightarrow4x^2-4x\left(y-5\right)+\left(y-5\right)^2+\left(y+2\right)^2=0\)
\(\Leftrightarrow\left(2x-y+5\right)^2+\left(y+2\right)^2=0\)
\(\Leftrightarrow\int^{2x-y+5=0}_{y+2=0}\Leftrightarrow\int^{x=\frac{-7}{2}}_{y=-2}\) loại vì x, y nguyên
vậy phương trình đã cho không có nghiệm nguyên
\(A=-x^2-4xy-5y^2-6y+1672\)
\(=-x^2-4xy-4y^2-y^2-6y-9+1681\)
\(=-\left(x^2+4xy+4y^2\right)-\left(y^2+6y+9\right)+1681\)
\(=-\left(x+2y\right)^2-\left(y+3\right)^2+1681\)
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 \(\left\{{}\begin{matrix}-\left(x+2y\right)^2=0\\-\left(y+3\right)^2=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x+2y=0\\y+3=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=-2y\\y=-3\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=6\\y=-3\end{matrix}\right.\)