NK
3
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
17 tháng 7 2015
=(x2-10xy+25y2)+(y2-6y+9)+14(x-5y)+49+1=[(x-5y)2+14(x-5y)+49]+(y-3)2+1=(x-5y+7)2+(y-3)2+1>=1
min=1khi y=3;x=8
TH
1
21 tháng 7 2015
A = [(x2 - 10xy + 25y2) + 2.(x - 5y).7 + 49 ] + (y2 - 6y + 9) + 1
= [(x -5y)2 + 2.(x - 5y) + 72] + (y - 3)2 + 1 = (x - 5y + 7)2 + (y - 3)2 + 1 \(\ge\) 0 + 0 + 1 = 1
=> GTNN của A bằng 1 khi x - 5y + 7 = 0 và y - 3 = 0
=> y = 3 và x = 8
B = (x2 + xy + \(\frac{y^2}{4}\)) - 2.(x + \(\frac{y}{2}\)). \(\frac{3}{2}\) + \(\frac{9}{4}\) + \(\frac{3y^2}{4}\) - \(\frac{3y}{2}\) + \(\frac{8023}{4}\)=[ (x + \(\frac{y}{2}\))2 - 2.(x + \(\frac{y}{2}\)). \(\frac{3}{2}\) + (\(\frac{3}{2}\))2 ] + 3. (\(\frac{y}{2}\) - 2)2 + \(\frac{7975}{4}\)
= (x + \(\frac{y}{2}\) - \(\frac{3}{2}\) )2 + 3. (\(\frac{y}{2}\) - 2)2 + \(\frac{7975}{4}\) \(\ge\) 0 + 0 + \(\frac{7975}{4}\) = \(\frac{7975}{4}\)
=> GTNN của B = \(\frac{7975}{4}\) khi x + \(\frac{y}{2}\) - \(\frac{3}{2}\) = 0 và \(\frac{y}{2}\) - 2 = 0
=> y = 4 và x = -1/2
VD
15 tháng 10 2020
Bài 1:
a)\(F=x^2+26y^2-10xy+14x-76y+59\)
\(=\left(x^2-2\cdot x\cdot5y+25y^2\right)+\left(14x-70y\right)+\left(y^2-6x+9\right)+50\)
\(=[\left(x-5y\right)^2+14\left(x-5y\right)+49]+\left(y-3\right)^2+1\)
\(=\left(x-5y+7\right)^2+\left(y-3\right)^2+1\ge1\)
Để Fmin=1 thì y=3;x=8
b)\(H=m^2-4mp+5p^2+10m-22p+28\)
\(=\left(m^2-2\cdot m\cdot2p+4p^2\right)+\left(10m-20p\right)+\left(p^2-2p+1\right)+27\)
\(=[\left(m-2p\right)^2+2\cdot\left(m-2p\right)\cdot5+25]+\left(p-1\right)^2+2\)
\(=\left(m-2p+5\right)^2+\left(p-1\right)^2+2\ge2\)
Để Hmin=2 thì p=1;m=-3
AN
0
25 tháng 8 2017
a, \(A_{\left(x\right)}=2x^2+2xy+y^2-2x+2y+2\)
\(=\left(x^2+y^2+1+2xy+2x+2y\right)+\left(x^2-4x+4\right)-3\)
\(=\left(x+y+1\right)^2+\left(x-2\right)^2-3\ge-3\) hay \(A_{\left(x\right)}\ge-3\)
Dấu ''='' xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y+1\right)^2=0\\\left(x-2\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x+y+1=0\\x-2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y=-3\\x=2\end{matrix}\right.\)
Vậy \(minA_{\left(x\right)}=-3\) khi x=-3; y=2
b, \(B_{\left(x\right)}=x^2-4xy+5y^2+10x-22y+28\)
\(=\left(x^2+4y^2+25-4xy+10x-20y\right)+\left(y^2-2y+1\right)+2\)
\(=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\Leftrightarrow B_{\left(x\right)}\ge2\)
Dấu ''='' xảy ra khi \(\left\{{}\begin{matrix}\left(x-2y+5\right)^2=0\\\left(y-1\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-2y+5=0\\y-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-3\\y=1\end{matrix}\right.\)
Vậy \(minB_{\left(x\right)}=2\Leftrightarrow x=-3;y=1\)
c, \(C_{\left(x\right)}=x^2-10xy+26y^2+14x-76y+59\)
\(=\left(x^2+25y^2+49-10xy+14x-70y\right)+\left(y^2-6y+9\right)+1\)
\(=\left(x-5y+7\right)^2+\left(y-3\right)^2+1\ge1\Leftrightarrow C_{\left(x\right)}\ge1\)
Dấu ''='' xảy ra khi \(\left\{{}\begin{matrix}\left(x-5y+7\right)^2=0\\\left(y-3\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-5y+7=0\\y-3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=8\\y=3\end{matrix}\right.\)
Vậy \(minC_{\left(x\right)}=1\Leftrightarrow x=8;y=3\)
d, \(D_{\left(x\right)}=4x^2-4xy+2y^2-20x-4y+174\)
\(=\left(4x^2+y^2+25-4xy-20x+10y\right)+\left(y-14y+49\right)+74\)
\(=\left(2x-y-5\right)^2+\left(y-7\right)^2+74\ge74\Leftrightarrow D_{\left(x\right)}\ge74\)
Dấu ''='' xảy ra khi \(\left\{{}\begin{matrix}\left(2x-y-5\right)^2=0\\\left(y-7\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x-y-5=0\\y-7=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=6\\y=7\end{matrix}\right.\)
Vậy \(minD_{\left(x\right)}=74\Leftrightarrow x=6;y=7\)
e, \(E_{\left(x\right)}=x^2-2x+y^2+4y+5\)
\(=\left(x^2-2x+1\right)+\left(y^2+4y+4\right)=\left(x-1\right)^2+\left(y+2\right)^2\ge0\)
Dấu ''='' xảy ra khi \(\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(y+2\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-1=0\\y+2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
Vậy \(minE_{\left(x\right)}=0\Leftrightarrow x=1;y=-2\)
25 tháng 8 2017
bạn ơi! Sao cái chỗ A(x) =(x+y+1)2+(x-2)2-3 mà chuyển sang lại là -3 v
DN
0
M = x2 + 26y2 - 10xy + 14x - 76y + 59
= ( x2 - 10xy + 25y2 + 14x - 70y + 49 ) + ( y2 - 6y + 9 ) + 1
= ( x - 5y + 7 )2 + ( y - 3 )2 + 1
Vì \(\hept{\begin{cases}\left(x-5y+7\right)^2\\\left(y-3\right)^2\end{cases}}\ge0\forall x,y\Rightarrow\left(x-5y+7\right)^2+\left(y-3\right)^2+1\ge1\forall x,y\)
Dấu "=" xảy ra khi x = 8 ; y = 3
Vậy MinM = 1 <=> x = 8 . y = 3
Ta có : \(M=x^2+26y^2-10xy+14x-76y+59\)
\(=\left(x^2-10xy+25y^2\right)+14\left(x-5y\right)+49+\left(y^2-6y+9\right)+1\)
\(=\left(x-5y\right)^2+14\left(x-5y\right)+49+\left(y-3\right)^2+1\)
\(=\left(x-5y+7\right)^2+\left(y-3\right)^2+1\ge1\forall x,y\)
Dấu \("="\)xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x-5y+7\right)^2=0\\\left(y-3\right)^2=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x-5y+7=0\\y-3=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x-15+7=0\\y=3\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=8\\y=3\end{cases}}\)
Vậy \(MinM=1\Leftrightarrow\hept{\begin{cases}x=8\\y=3\end{cases}}\)