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\(C=x^2-xy+y^2-2x-2y\Leftrightarrow2C=2x^2-2xy+2y^2-4x-4y\)
\(\Leftrightarrow2C=\left(x^2-2xy+y^2\right)+\left(x^2-4x+4\right)+\left(y^2-4y+4\right)-8\)
\(\Leftrightarrow2C=\left(x-y\right)^2+\left(x-2\right)^2+\left(y-2\right)^2-8\)
\(\Leftrightarrow C=\frac{\left(x-y\right)^2+\left(x-2\right)^2+\left(y-2\right)^2-8}{2}\)
\(\Leftrightarrow C=\frac{\left(x-y\right)^2+\left(x-2\right)^2+\left(y-2\right)^2}{2}-4\)
Vì \(\frac{\left(x-y\right)^2+\left(x-2\right)^2+\left(y-2\right)^2}{2}\ge0\)
\(\Rightarrow\)\(\frac{\left(x-y\right)^2+\left(x-2\right)^2+\left(y-2\right)^2}{2}-4\ge-4\)
Hay \(C\ge-4\)
Vậy \(GTNN\) của \(C=-4\Leftrightarrow\hept{\begin{cases}x-y=0\\x-2=0\\y-2=0\end{cases}}\Leftrightarrow x=y=2\)
2A = 2x^2 - 2xy + 2y^2 - 4x - 4y
2A = ( x^2 - 2xy + y^2 ) + ( x^2 - 4x + 2^2 ) + ( y^2 - 4y + 2^2 ) - 8
2A = ( x - y )^2 + ( x - 2 )^2 + ( y - 2 )^2 - 8
Ta có : ( x - y )^2 >= 0 ; ( x - 2 )^2 >= 0 ; ( y - 2 )^2 >= 0 với mọi x , y
=> Min 2A = 0 + 0 + 0 - 8 = -8
=> Min A = -8 : 2 = -4
a, \(A=x^4-2x^3+2x^2-2x+3\)
\(=\left(x^4+2x^2+1\right)-\left(2x^3+2x\right)+2\)
\(=\left(x^2+1\right)^2-2x\left(x^2+1\right)+2\)
\(=\left(x^2+1\right)\left(x^2-2x+1\right)+2\)
\(=\left(x^2+1\right)\left(x-1\right)^2+2\)
Vì \(\hept{\begin{cases}x^2\ge0\\\left(x-1\right)^2\ge0\end{cases}\Rightarrow\hept{\begin{cases}x^2+1\ge1\\\left(x-1\right)^2\ge0\end{cases}\Rightarrow}\left(x^2+1\right)\left(x-1\right)^2\ge0}\)
\(\Rightarrow A=\left(x^2+1\right)\left(x-1\right)^2+2\ge2\)
Dấu "=" xảy ra khi x = 1
Vậy Amin = 2 khi x = 1
b, \(B=4x^2-2\left|2x-1\right|-4x+5=\left(4x^2-4x+1\right)-2\left|2x-1\right|+4=\left(2x-1\right)^2-2\left|2x-1\right|+4\)
đề sai ko
c, \(C=4-x^2+2x=-\left(x^2-2x+1\right)+5=-\left(x-1\right)^2+5\)
Vì \(-\left(x-1\right)^2\le0\Rightarrow C=-\left(x-1\right)^2+5\le5\)
Dấu "=" xảy ra khi x=1
Vậy Cmin = 5 khi x = 1
2/
+) \(D=-x^2-y^2+x+y+3=-\left(x^2-x+\frac{1}{4}\right)-\left(y^2-y+\frac{1}{4}\right)+\frac{7}{2}=-\left(x-\frac{1}{2}\right)^2-\left(y-\frac{1}{2}\right)^2+\frac{7}{2}\)
Vì \(\hept{\begin{cases}-\left(x-\frac{1}{2}\right)^2\le0\\-\left(y-\frac{1}{2}\right)^2\le0\end{cases}\Rightarrow-\left(x-\frac{1}{2}\right)^2-\left(y-\frac{1}{2}\right)^2\le0}\Rightarrow D=-\left(x-\frac{1}{2}\right)^2-\left(y-\frac{1}{2}\right)^2+\frac{7}{2}\le\frac{7}{2}\)
Dấu "=" xảy ra khi x=y=1/2
Vậy Dmax=7/2 khi x=y=1/2
+) Đề sai
+)bài này là tìm min
\(G=x^2-3x+5=\left(x^2-3x+\frac{9}{4}\right)+\frac{11}{4}=\left(x-\frac{3}{2}\right)^2+\frac{11}{4}\ge\frac{11}{4}\)
Dấu "=" xảy ra khi x=3/2
Vậy Gmin=11/4 khi x=3//2
1) \(\left(3x+2\right)^2-\left(x-6\right)^2=\left(3x+2-x+6\right)\left(3x+2+x-6\right)=\left(2x+8\right)\left(4x-4\right)=8\left(x+4\right)\left(x-1\right)\)
2) \(A=x^2+2y^2+2xy-2y+2021=\left(x^2+2xy+y^2\right)+\left(y^2-2y+1\right)+2020=\left(x+y\right)^2+\left(y-1\right)^2+2020\ge2020\)
\(minA=2020\Leftrightarrow\)\(\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)
\(A=2x^2+y^2-2xy-2x+3\)
\(A=\left(x^2-2xy+y^2\right)+\left(x^2-2x+1\right)+2\)
\(A=\left(x-y\right)^2+\left(x-1\right)^2+2\)
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
D= x^2+2*(1/2)xy+((1/2)y)^2+(3/4)y^2+1
=(x+(1/2)y)^2 +1
Nên min D=1
E=(2x-1)^2+(y-1)^2+(x-3y)^2+1
nên min E=1
Lời giải:
Ta có:
$2P=2x^2-2xy+2y^2-4x-4y=(x^2-2xy+y^2)+(x^2-4x+4)+(y^2-4y+4)-8$
$=(x-y)^2+(x-2)^2+(y-2)^2-8\geq 0+0+0-8=-8$
$\Rightarrow P\geq -4$
Vậy $P_{\min}=-4$. Giá trị này đạt được khi $x-y=x-2=y-2=0$
$\Leftrightarrow x=y=2$