Đang onl bằng điện thoại nên mình làm sơ sơ thôi nhé :((
A = ( x2 - 3x + 9/4 ) + ( y2 - 4y + 4 ) - 5/4
= ( x - 3/2 )2 + ( y - 2 )2 - 5/4 >= -5/4
Dấu = xảy ra <=> x = 3/2 ; y = 2
Vậy ...
B = ( x2 - 2xy + y2 ) + ( y2 + 4y + 4 ) - 11
= ( x - y )2 + ( y + 2 )2 - 11 >= -11
Dấu = xảy ra <=> x = y = -2
Vậy ...
a) \(A=x^2+4y^2-3x-4y+5\)
\(=\left(x^2-3x+\frac{9}{4}\right)+\left(4y^2-4y+1\right)+\frac{7}{4}\)
\(=\left(x-\frac{3}{2}\right)^2+\left(2y-1\right)^2+\frac{7}{4}\)
Vì \(\left(x-\frac{3}{2}\right)^2\ge0\forall x\); \(\left(2y-1\right)^2\ge0\forall y\)
\(\Rightarrow\left(x-\frac{3}{2}\right)^2+\left(2y-1\right)^2\ge0\forall x,y\)
\(\Rightarrow\left(x-\frac{3}{2}\right)^2+\left(2y-1\right)^2+\frac{7}{4}\ge\frac{7}{4}\forall x,y\)
Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}x-\frac{3}{2}=0\\2y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{3}{2}\\2y=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{3}{2}\\y=\frac{1}{2}\end{cases}}\)
Vậy \(minA=\frac{7}{4}\)\(\Leftrightarrow\hept{\begin{cases}x=\frac{3}{2}\\y=\frac{1}{2}\end{cases}}\)
b) \(B=x^2+2y^2-2xy+4y-7\)
\(=\left(x^2-2xy+y^2\right)+\left(y^2+4y+4\right)-11\)
\(=\left(x-y\right)^2+\left(y+2\right)^2-11\)
Vì \(\left(x-y\right)^2\ge0\forall x,y\); \(\left(y+2\right)^2\ge0\forall y\)
\(\Rightarrow\left(x-y\right)^2+\left(y+2\right)^2\ge0\forall x,y\)
\(\Rightarrow\left(x-y\right)^2+\left(y+2\right)^2-11\ge-11\forall x,y\)
Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}x-y=0\\y+2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=y\\y=-2\end{cases}}\Leftrightarrow x=y=-2\)
Vậy \(minB=-11\)\(\Leftrightarrow x=y=-2\)