a: A=x^2-2xy+y^2+y^2-4y+4+1

=(x-y)^2+(y-2)^2+1>=1
Dấu = xảy ra khi x=y=2

b: B=4x^2+8xy+4y^2+x^2-2x+1+y^2+2y+1-2

=(2x+2y)^2+(x-1)^2+(y+1)^2-2>=-2

Dấu = xảy ra khi x=1 và y=-1

\(K=x^2-2xy+2y^2-4y+2016=\)\(x^2-2xy+y^2+y^2-4y+4+2012=\)\(\left(x^2-2xy+y^2\right)+\left(y^2-4y+4\right)+2012=\)\(\left(x-y\right)^2+\left(y-2\right)^2+2012\)

Vì \(\left(x-y\right)^2\ge0;\left(y-2\right)^2\ge0\)

\(\Rightarrow K_{min}=2012\) Khi \(\hept{\begin{cases}\left(x-y\right)^2=0\\\left(y-2\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x-y=0\\y-2=0\end{cases}\Rightarrow}\hept{\begin{cases}x=y\\y=2\end{cases}\Rightarrow}x=y=2}\)

\(x^2-2xy+2y^2-4y+2016\)

\(\Leftrightarrow x^2-2xy+y^2+y^2-4y+4+2012\)

\(\Leftrightarrow\left(x-y\right)^2+\left(y-2\right)^2+2014\)

Xét đa thức \(\left(x-y\right)^2+\left(y-2\right)^2\)

Dễ thấy \(\left(x-y\right)^2+\left(y-2\right)^2\) luôn luôn dương với mọi giá trị của \(x,y\)

Vậy giá trị nhỏ nhất của k=2014

a, \(2x^2-4xy+4y^2-6x\)

\(=x^2-2xy-2xy+4y^2+x^2-3x-3x+9-9\)

\(=\left(x-2y\right)^2+\left(x-3\right)^2-9\)

Với mọi giá trị của \(x;y\in R\) ta có:

\(\left(x-2y\right)^2+\left(x-3\right)^2-9\ge-9\)

Để \(\left(x-2y\right)^2+\left(x-3\right)^2-9=-9\) thì

\(\left\{{}\begin{matrix}\left(x-2y\right)^2=0\\\left(x-3\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3-2y=0\\x=3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y=1,5\\x=3\end{matrix}\right.\)

Vậy..............

b, \(z^2-4zt+5t^2-2t+13\)

\(=z^2-2zt-2zt+4t^2+t^2-t-t+1+12\)

\(=\left(z-2t\right)^2+\left(t-1\right)^2+12\)

Với mọi giá trị của \(z;t\in R\) ta có:

\(\left(z-2t\right)^2+\left(t-1\right)^2+12\ge12\)

Để \(\left(z-2t\right)^2+\left(t-1\right)^2+12=12\) thì

\(\left\{{}\begin{matrix}\left(z-2t\right)^2=0\\\left(t-1\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}z-2=0\\t=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}z=2\\t=1\end{matrix}\right.\)

Vậy...............

Câu c tường tự !!!

a,Đặt A= \(2x^2-4xy+4y^2-6x\)

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

\(=2\left(x^2-2xy-3x\right)+4y^2\)

\(=2\left[x^2-2x\left(y+\dfrac{3}{2}\right)+\left(y+\dfrac{3}{2}\right)^2\right]+4y^2-\left(y+\dfrac{3}{2}\right)^2\)

\(=2\left(x-y-\dfrac{3}{2}\right)^2+4y^2-y^2-3y-\dfrac{9}{4}\)

\(=2\left(x-y-\dfrac{3}{2}\right)^2+3\left(y^2-y+\dfrac{1}{4}\right)-3\)

\(=2\left(x-y-\dfrac{3}{2}\right)^2+3\left(y-\dfrac{1}{2}\right)^2-3\)

Với mọi giá trị của x;y ta có:

\(\left(x-y-\dfrac{3}{2}\right)^2\ge0;\left(y-\dfrac{1}{2}\right)^2\ge0\)

\(\Rightarrow2\left(x-y-\dfrac{3}{2}\right)^2+\left(y-\dfrac{1}{2}\right)^2-3\ge-3\)

Vậy Min A = -3 khi \(\left\{{}\begin{matrix}x-y-\dfrac{3}{2}=0\\y-\dfrac{1}{2}=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x-\dfrac{1}{2}-\dfrac{3}{2}=0\\y=\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x-2=0\\y=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=\dfrac{1}{2}\end{matrix}\right.\)

b, Đặt B = \(z^2-4zt+5t^2-2t+13\)

\(=\left(z^2-4zt+4t^2\right)+\left(t^2-2t+1\right)+12\)

\(=\left(z-2t\right)^2+\left(t-1\right)^2+12\)

Với mọi giá trị của z;t ta có:

\(\left(z-2t\right)^2\ge0;\left(t-1\right)^2\ge0\)

\(\Rightarrow\left(z-2t\right)^2+\left(t-1\right)^2+12\ge12\)

Vậy Min B = 12 khi \(\left\{{}\begin{matrix}z-2t=0\\t-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}z-2=0\\t=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}z=2\\t=1\end{matrix}\right.\)

c, Đặt C = \(16x^2-8x+y^2-2y\)

\(=\left(16x^2-8x+1\right)+\left(y^2-2y+1\right)-2\)

\(=\left(4x-1\right)^2+\left(y-1\right)^2-2\)

Với mọi giá trị x;y ta có:

\(\left(4x-1\right)^2\ge0;\left(y-1\right)^2\ge0\)

\(\Rightarrow\left(4x-1\right)^2+\left(y-1\right)^2-2\ge-2\)

Vậy Min C = -2 khi \(\left\{{}\begin{matrix}4x-1=0\\y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x=1\\y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{4}\\y=1\end{matrix}\right.\)

\(K=\left(x^2-2xy+y^2\right)+\left(y^2-4y+4\right)+2012=\left(x-y\right)^2+\left(y-2\right)^2+2012\ge2012\)Min K = 2012 <=> x = y = 2

nhanh lên các bạn nhé mai mình đi học rồi

a) x²+y²-4x+4y+8=0

⇔ (x-2)²+(y+2)²=0

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

b)5x²-4xy+y²=0

⇔ x²+(2x-y)²=0

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

c)x²+2y²+z²-2xy-2y-4z+5=0

⇔ (x-y)²+(y-1)²+(z-2)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y-1=0\\z-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y=1\\z=2\end{matrix}\right.\)

b: Ta có: \(5x^2-4xy+y^2=0\)

\(\Leftrightarrow x^2-\dfrac{4}{5}xy+y^2=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{2}{5}y+\dfrac{4}{25}y^2+\dfrac{21}{25}y^2=0\)

\(\Leftrightarrow\left(x-\dfrac{2}{5}y\right)^2+\dfrac{21}{25}y^2=0\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)

d)3x²+3y²+3xy-3x+3y+3=0

⇔ 6x²+6y²+6xy-6x+6y+6=0

⇔ 3(x+y)²+3(x-1)²+3(y+1)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\x-1=0\\y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)