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a/ \(M=x^2+y^2-x+6y+10=\left(x^2-x+\frac{1}{4}\right)+\left(y^2+6y+9\right)+10-\frac{1}{4}-9\)
\(=\left(x-\frac{1}{2}\right)^2+\left(y+3\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Suy ra Min M = 3/4 <=> (x;y) = (1/2;-3)
b/
1/ \(A=4x-x^2+3=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\)
Suy ra Min A = 7 <=> x = 2
2/ \(B=x-x^2=-\left(x^2-x+\frac{1}{4}\right)+\frac{1}{4}=-\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)
Suy ra Min B = 1/4 <=> x = 1/2
3/ \(N=2x-2x^2-5=-2\left(x^2-x+\frac{1}{4}\right)-5+\frac{1}{2}=-2\left(x-\frac{1}{2}\right)^2-\frac{9}{2}\)
\(\ge-\frac{9}{2}\)
Suy ra Min N = -9/2 <=> x = 1/2
Ta có: \(x^2-2x+y^2-4y+7\)
\(=\left(x^2-2x+1\right)+\left(y^2-4y+4\right)+2\)
\(=\left(x-1\right)^2+\left(y-2\right)^2+2\)
Vì:\(\left(x-1\right)^2+\left(y-2\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-1\right)^2+\left(y-2\right)^2+2\ge2\forall x\)
Dấu = xảy ra khi: \(\hept{\begin{cases}\left(x-1\right)^2=0\\\left(y-2\right)^2=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=1\\y=2\end{cases}}\)
Vậy:GTNN của bt là 2 tại x=1,y=2
\(M=\left(x^2-x+\frac{1}{4}\right)+\left(y^2-6y+9\right)+\frac{3}{4}=\left(x-\frac{1}{2}\right)^2+\left(y-3\right)^2+\frac{3}{4}\ge\frac{3}{4}\Rightarrow MinM=\frac{3}{4}\Leftrightarrow x=\frac{1}{2};y=3\)\(P=x^2-2x+1+4=\left(x-1\right)^2+4\ge4\Rightarrow MinP=4\Leftrightarrow x=1\)
Câu b mình viết nhầm dấu \(\ge\)đáng lẽ đúng phải là \(\le\)
a)
\(A=x^2+y^2-x+6y+10.\)
\(=\left(x^2-x+\frac{1}{4}\right)+\left(y^2+6y+9\right)+\frac{3}{4}\)
\(=\left(x-\frac{1}{2}\right)^2+\left(y+3\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Vậy \(MinA=\frac{3}{4}\Leftrightarrow\hept{\begin{cases}\left(x-\frac{1}{2}\right)^2=0\\\left(y+3\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x-\frac{1}{2}=0\\y+3=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{1}{2}\\y=-3\end{cases}}}\)
b)
\(B=2x-2x^2-5\)
\(=-2\left(x^2-x+\frac{1}{4}\right)+2.\frac{1}{4}-5\)
\(=-2\left(x-\frac{1}{2}\right)^2-\frac{9}{2}\ge-\frac{9}{2}\)
Vậy \(MaxB=-\frac{9}{2}\Leftrightarrow\left(x-\frac{1}{2}\right)^2=0\Leftrightarrow x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{2}\)