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a, B=x2+4xy+y2+x2-8x+16+2012
B=(x+y) 2+(x-4)2+2012
Vậy B >=2012 ( Dấu "=" xảy ra khi x=4,y=-4)
b làm tương tự
c, 9x2+6x+1+y2-4y+4+x2-4xz+4z2=0
(3x+1)2+(y-4)2+(x-2z)2=0
Vậy 3x+1=0 => x = -1/3
y-4=0 => y=4
x-2z=0 thế x=-1/3 ta được. -1/3-2z=0 => z = -1/6
Bạn nhớ ghi lại đề minh không ghi đề
a) \(B=2x^2+y^2+2xy-8x+2028\)
\(=\left(x^2+2xy+y^2\right)+\left(x^2-8x+4^2\right)+2012=\left(x+y\right)^2+\left(x-4\right)^2+2012\ge2012\)
\(MinB=2012\Leftrightarrow\hept{\begin{cases}x=4\\y=-4\end{cases}}\)
b)\(C=x^2+5y^2+4xy+2x+2y-7\)
\(=\left(x^2+4xy+4y^2\right)+\left(2x+4y\right)+1+\left(y^2-2y+1\right)-9\)
\(=\left(\left(x+2y\right)^2+2\left(x+2y\right)+1\right)+\left(y-1\right)^2-9=\left(x+2y+1\right)^2+\left(y-1\right)^2-9\ge9\)
\(MinC=-9\Leftrightarrow\hept{\begin{cases}x+2y+1=0\\y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-3\\y=1\end{cases}}\)
c)\(10x^2+y^2+4z^2+6x-4y-4xz+5=0\)
\(\Leftrightarrow\left(9x^2+6x+1\right)+\left(y^2-4y+4\right)+\left(x^2-4xz+4z^2\right)=0\)
\(\Leftrightarrow\left(3x+1\right)^2+\left(y-2\right)^2+\left(x-2z\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}3x+1=0\\y-2=0\\x-2z=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-\frac{1}{3}\\y=2\\z=-\frac{1}{6}\end{cases}}\)
Ta có:
D=2x2+3y2+4xy−8x−2y+18C=2x2+3y2+4xy−8x−2y+18
D=2(x2+2xy+y2)+y2−8x−2y+18C=2(x2+2xy+y2)+y2−8x−2y+18
D=2[(x+y)2−4(x+y)+4]+(y2+6y+9)+1C=2[(x+y)2−4(x+y)+4]+(y2+6y+9)+1
D=2(x+y−2)2+(y+3)2+1≥1C=2(x+y−2)2+(y+3)2+1≥1
Dấu "=" xảy ra ⇔x+y=2⇔x+y=2và y=−3y=−3
Hay x = 5 , y = -3
Đc chx bạn
a/ \(=\left(9x^2+30x+25\right)+\left(x^2+10x+25\right)=\)
\(=\left(3x+5\right)^2+\left(x+5\right)^2\)
b/ \(=\left(16x^2+8x+1\right)+\left(y^2-4y+4\right)=\left(4x+1\right)^2+\left(y-2\right)^2\)
c/
\(A=x^2-4xy+2x-4y+3+4y^2\)
\(A=x^2-2.2xy+\left(2y\right)^2+2x-4y+3\)
\(A=\left(x-2y\right)^2-2.\left(x-2y\right)+1+2\)
\(A=\left(x-2y-1\right)^2+2\ge2\)
Vậy GTNN của A=2.
\(x^2-2x-3=0\)
\(\Leftrightarrow x^2+x-3x-3=0\)
\(\Leftrightarrow x\left(x+1\right)-3\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\)
Vậy x = -1 hoặc x = 3
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 \(\left\{{}\begin{matrix}4x-1=0\\y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x=1\\y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{4}\\y=1\end{matrix}\right.\)