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\(A=2x^2+y^2-2xy-2x+3\)
\(A=\left(x^2-2xy+y^2\right)+\left(x^2-2x+1\right)+2\)
\(A=\left(x-y\right)^2+\left(x-1\right)^2+2\)
Mà \(\left(x-y\right)^2\ge0\forall x;y\)
\(\left(x-1\right)^2\ge0\forall x\)
\(\Rightarrow A\ge2\)
Dấu "=" xảy ra khi :
\(\hept{\begin{cases}x-y=0\\x-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}y=1\\x=1\end{cases}}\)
Vậy Min A = 2 khi x=y=1
\(4x^2+y^2-2xy-2x+2y=\left(x^2+y^2+1-2xy-2x+2y\right)+3x^2.\)
\(=\left(x-y-1\right)^2+3x^2\ge0\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}\left(x-y-1\right)^2=0\\3x^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0\\y=-1\end{cases}}\)
\(C=2x^2+9y^2-6xy-2x+2018\)
\(=\left(x^2-6xy+9y^2\right)+\left(x^2-2x+1\right)+2017\)
\(=\left(x-3y\right)^2+\left(x-1\right)^2+2017\)
Nhận xét :
\(\left\{{}\begin{matrix}\left(x-3y\right)^2\ge0\\\left(x-1\right)^2\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left(x-3y\right)^2+\left(x-1\right)^2\ge0\)
\(\Leftrightarrow\left(x-3y\right)^2+\left(x-1\right)^2+2017\ge2017\)
\(\Leftrightarrow C\ge2017\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left(x-3y\right)^2=0\\\left(x-1\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{1}{3}\end{matrix}\right.\)
Vậy \(C_{Min}=2017\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{1}{3}\end{matrix}\right.\)
\(D=x^2-2xy+6y^2-12x+2y+45\)
\(=\left(x^2-2xy+y^2\right)-\left(12x+12y\right)-10y+5y^2+45\)
\(=\left(x-y\right)^2-12\left(x-y\right)+36+\left(5y^2-10y+5\right)+4\)
\(=\left(x-y-6\right)^2+5\left(y-1\right)^2+4\)
Nhận xét :
\(\left\{{}\begin{matrix}\left(x-y-6\right)^2\ge0\\5\left(y-1\right)^2\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left(x-y-6\right)+5\left(y-1\right)^2+4\ge4\)
\(\Leftrightarrow D\ge4\)
Dấu "=" xảy ra khi \(\Leftrightarrow\left\{{}\begin{matrix}x=7\\y=1\end{matrix}\right.\)
Vậy \(D_{Min}=4\Leftrightarrow\left\{{}\begin{matrix}x=7\\y=1\end{matrix}\right.\)
a) \(A=2x^2+9y^2-6xy-6x-12y+2014\)
\(=\left(2x^2-6xy-6x\right)+\left(9y^2-12y\right)+2014\)
\(=2\left[x^2-2.x.\frac{3\left(y+1\right)}{2}+\frac{9\left(y+1\right)^2}{4}\right]+\left[9y^2-12y-\frac{9}{2}.\left(y+1\right)^2\right]+2014\)
\(=2\left[x-\frac{3\left(y+1\right)}{2}\right]^2+\frac{1}{2}\left(3y-7\right)^2+1985\ge1985\)
Dấu "=" xảy ra khi và chỉ khi y = \(\frac{7}{3}\Rightarrow x=5\)
Vậy Min A = 1985 tại \(\left(x;y\right)=\left(5;\frac{7}{3}\right)\)
b) \(B=-x^2+2xy-4y^2+2x+10y-8\)
\(=-\left(x^2-2xy-2x\right)-\left(4y^2-10y\right)-8\)
\(=-\left[x^2-2x\left(y+1\right)+\left(y+1\right)^2\right]-\left[4y^2-10y-\left(y+1\right)^2\right]-8\)
\(=-\left(x-y-1\right)^2-\left(y-2\right)^2+5\le5\)
Dấu đẳng thức xảy ra khi và chỉ khi y = 2 => x = 3
Vậy B đạt giá trị lớn nhất bằng 5 tại (x;y) = (3;2)
bạn ghi sai đề rồi phải là - 2y mới làm dc nha
\(B=x^2+2x\left(y-1\right)+\left(y-1\right)^2+x^2+1\)
\(B=\left(x+y-1\right)^2+x^2+1\ge1\)
Min của B = 1 khi x+y -1=0,x^2=0
=> x+y= 1 , x=0
=> x=0,y=1
Ung hộ mình nha
a,Đặt A= \(2x^2+2xy+y^2-2x+2y+15\)
\(=\left(x^2+y^2+1+2xy+2x+2y\right)+\left(x^2-4x+4\right)+10\)
\(=\left(x+y+1\right)^2+\left(x-2\right)^2+10\)
Vì \(\left(x+y+1\right)^2\ge0;\left(x-2\right)^2\ge0\Rightarrow\left(x+y+1\right)^2+\left(x-2\right)^2+10\ge0\)
hay \(A\ge10\)
Dấu ''='' xảy ra khi \(\left\{{}\begin{matrix}\left(x+y+1\right)^2=0\\\left(x-2\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x+y+1=0\\x=2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y=-3\\x=2\end{matrix}\right.\)
Vậy min A=10 khi x=2; y=-3
b/ \(=\left(x^2-2xy+y^2\right)+\left(3x^2-12x+12\right)+\left(8y^2-32y+32\right)-4\)
=\(\left(x-y\right)^2+3\left(x-2\right)^2+8\left(y-2\right)^2-4\ge-4\)
Vậy Min =-4 khi x=y=2