dạt dấu trừ ra ngoài

mơn bb*

\(A=-x^2-4xy-5y^2+6y+1672\)

\(A=-x^2-4xy-4y^2-y^2+6y-9+1681\)

\(A=-\left(x+2y\right)^2-\left(y-3\right)^2+1681\)

\(A=1681-\left[\left(x+2y\right)^2+\left(y-3\right)^2\right]\)

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 \(\hept{\begin{cases}x=-6\\y=3\end{cases}}\)

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...

Trần Thủy Dung làm sai rồi +6y=-(-6y) chứ

\(x^2+5y^2-4xy-6y+9=0\)

\(\Leftrightarrow\left(x^2-4xy+4y^2\right)+\left(y^2-6y+9\right)=0\)

\(\Leftrightarrow\left(x-2y\right)^2+\left(y-3\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-2y=0\\y-3=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=6\\y=3\end{matrix}\right.\)

\(\Rightarrow A=...\)

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

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