Tìm giá trị nhỏ nhất của biểu thức
A = \(x^2-xy+y^2-2x-2y\)
K
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N
0
NT
1
S
29 tháng 12 2020
A= x2+2y2-2xy-2x-2y+1015
A = x2 - 2xy - 2x + y2 + 2y + 1 + y2 - 4y + 4 + 1010
A = [x2 - 2x(y + 1) + (y+1)2 ] + (y-2)2 + 1010
A = ( x - y - 1)2 + (y-2)2 + 1010 \(\ge1010\forall x,y\)
Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}x-y-1=0\\y-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\)
Vậy MinA = 1010 <=> \(\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\)
19 tháng 5 2021
\(gt\Leftrightarrow\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}=1\)
\(P=\dfrac{1}{xyz}\left(x\sqrt{2y^2+yz+2z^2}+y\sqrt{2x^2+xz+2z^2}+z\sqrt{2y^2+xy+2x^2}\right)\)
\(=\dfrac{1}{xyz}\left(x\sqrt{\dfrac{5}{4}\left(y+z\right)^2+\dfrac{3}{4}\left(y-z\right)^2}+y\sqrt{\dfrac{5}{4}\left(x+z\right)^2+\dfrac{3}{4}\left(x-z\right)^2}+z\sqrt{\dfrac{5}{4}\left(x+y\right)^2+\dfrac{3}{4}\left(x-y\right)^2}\right)\)
\(\ge\dfrac{1}{xyz}\left[x.\dfrac{\sqrt{5}\left(z+y\right)}{2}+y.\dfrac{\sqrt{5}\left(x+z\right)}{2}+z.\dfrac{\sqrt{5}\left(x+y\right)}{2}\right]\)
\(=\dfrac{\sqrt{5}\left(z+y\right)}{2yz}+\dfrac{\sqrt{5}\left(x+z\right)}{2xz}+\dfrac{\sqrt{5}\left(x+y\right)}{2xy}\)
\(=\dfrac{\sqrt{5}}{3}\left(1+1+1\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge\dfrac{\sqrt{5}}{3}\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}\right)^2=\dfrac{\sqrt{5}}{3}\) (bunhia)
Dấu = xảy ra khi \(x=y=z=9\)
KB
19 tháng 5 2021
Thấy : \(\sqrt{2y^2+yz+2z^2}=\sqrt{\dfrac{5}{4}\left(y+z\right)^2+\dfrac{3}{4}\left(y-z\right)^2}\ge\dfrac{\sqrt{5}}{2}\left(y+z\right)>0\)
CMTT : \(\sqrt{2x^2+xz+2z^2}\ge\dfrac{\sqrt{5}}{2}\left(x+z\right)\) ; \(\sqrt{2y^2+xy+2x^2}\ge\dfrac{\sqrt{5}}{2}\left(x+y\right)\)
Suy ra : \(P\ge\dfrac{1}{xyz}.\dfrac{\sqrt{5}}{2}\left[x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)\right]\)
\(\Rightarrow P\ge\sqrt{5}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)
Ta có : \(\sqrt{xy}+\sqrt{yz}+\sqrt{xz}=\sqrt{xyz}\Leftrightarrow\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}=1\)
Mặt khác : \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}\right)^2}{3}=\dfrac{1}{3}\)
Suy ra : \(P\ge\dfrac{\sqrt{5}}{3}\)
" = " \(\Leftrightarrow x=y=z=9\)
HN
2
B
3 tháng 9 2021
Từ gt ta có x^2+y^^2=xy+1
=>P=(x^2+y^2)^2-2x^2y^2-x^2y^2
=(xy+1)2-2x2y2-x2y2
=x2y2+xy+1-3x2y2=-2x2y2+xy+1
=......
NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 9 2021
\(1=x^2+y^2-xy\ge2xy-xy=xy\Rightarrow xy\le1\)
\(1=x^2+y^2-xy\ge-2xy-xy=-3xy\Rightarrow xy\ge-\dfrac{1}{3}\)
\(\Rightarrow-\dfrac{1}{3}\le xy\le1\)
\(P=\left(x^2+y^2\right)^2-2\left(xy\right)^2-\left(xy\right)^2=\left(xy+1\right)^2-3\left(xy\right)^2=-2\left(xy\right)^2+2xy+1\)
Đặt \(xy=t\in\left[-\dfrac{1}{3};1\right]\)
\(P=f\left(t\right)=-2t^2+2t+1\)
\(f'\left(t\right)=-4t+2=0\Rightarrow t=\dfrac{1}{2}\)
\(f\left(-\dfrac{1}{3}\right)=\dfrac{1}{9}\) ; \(f\left(\dfrac{1}{2}\right)=\dfrac{3}{2}\) ; \(f\left(1\right)=1\)
\(\Rightarrow P_{max}=\dfrac{3}{2}\) ; \(P_{min}=\dfrac{1}{9}\)
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