3N
5
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
LD
1
5 tháng 10 2017
(9x2-18x+9)+(y2-6y+9)+2(z2+2z+1)=0\(\Rightarrow\)(3x-3)2+(y-3)2+2(z+1)2=0\(\Rightarrow\hept{\begin{cases}\left(3x-3\right)^2=0\\\left(y-3\right)^2=0\\\left(z+1\right)^2=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=1\\y=3\\z=-1\end{cases}}\)
MN
0
16 tháng 2 2017
ta co 9(x^2-2x+1) +( y^2 -6y +9) + 2(z^2 + 2z +1) = 0
suy ra 9(x-1)^2 + (y - 3 )^2 + 3( z-1)^2 = 0
suy ra x-1=0 ; y-3 =0 ; z-1=0
suy ra x=1;y=3; z=1
VH
1
Y
7 tháng 8 2018
\(9x^2+y^2+2z^2-18x+4z-6y+20=0\)
\(\Rightarrow\left[\left(3x\right)^2-2.3x.3+9\right]+\left(y^2-2.y.3+9\right)+\left(2z^2+4z+2\right)=0\)
\(\Rightarrow\left(3x-3\right)^2+\left(y-3\right)^2+2\left(z^2+2z+1\right)=0\)
\(\Rightarrow\left(3x-3\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\)
Vì \(\left(3x-3\right)^2\ge0\) với mọi x
\(\left(y-3\right)^2\ge0\) với mọi y
\(2\left(z+1\right)^2\ge0\) với mọi z
\(\Rightarrow\left(3x-3\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2\ge0\) với mọi x, y, z
Mà \(\left(3x-3\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left(3x-3\right)^2=0\\\left(y-3\right)^2=0\\2\left(z+1\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3x-3=0\\y-3=0\\\left(z+1\right)^2=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}3\left(x-1\right)=0\\y=3\\z+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-1=0\\y=3\\z=-1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)
Vậy x = 1 ; y = 3 ; z = -1
14 tháng 12 2016
\(9x^2+y^2+2z^2-18x+4z-6y+20=0\)
\(\Leftrightarrow\left(9x^2-18x+9\right)+\left(y^2-6y+9\right)+\left(2z^2+4z+2\right)=0\)
\(\Leftrightarrow9\left(x^2-2x+1\right)+\left(y-3\right)^2+2\left(z^2+2z+1\right)=0\)
\(\Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\)
\(\Leftrightarrow\begin{cases}9\left(x-1\right)^2=0\\\left(y-3\right)^2=0\\2\left(z+1\right)^2=0\end{cases}\)\(\Leftrightarrow\begin{cases}x-1=0\\y-3=0\\z+1=0\end{cases}\)
\(\Leftrightarrow\begin{cases}x=1\\y=3\\z=-1\end{cases}\)
A
1
AT
30 tháng 3 2018
\(9x^2+y^2+2z^2-18x+4z-6y+20=0\)
\(\left(9x^2-18x+9\right)+\left(y^2-6y+9\right)+\left(2z^2+4z+2\right)=0\)
\(\Leftrightarrow\left(3x-3\right)^2+\left(y-2\right)^2+\left(\sqrt{2}z+\sqrt{2}\right)^2=0\)
Vì: \(\left(3x-3\right)^2+\left(y-2\right)^2+\left(\sqrt{2}z+\sqrt{2}\right)^2\ge0\forall x,y,z\)
=> Dấu = xảy ra khi: \(\left\{{}\begin{matrix}3x-3=0\\y-2=0\\\sqrt{2}z+\sqrt{2}=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\\z=-1\end{matrix}\right.\)
Vậy.................
\(9x^2+y^2+2z^2-18x+4z-6y+20=0\)
\(\Leftrightarrow9\left(x^2-2x+1\right)+\left(y^2-6y+9\right)+2\left(z^2+2z+1\right)=0\)
\(\Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}x-1=0\\y-3=0\\z+1=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=1\\y=3\\z=-1\end{cases}}\)
vậy......
Ta có: \(9x^2+y^2+2z^2-18x+4z-6y+20=0\)
\(\Leftrightarrow\left(9x^2-18x+9\right)+\left(y^2-6y+9\right)+\left(2z^2+4z+2\right)=0\)
\(\Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\)
Mà \(VT\ge0\left(\forall x,y,z\right)\) nên dấu "=" xảy ra khi:
\(\hept{\begin{cases}9\left(x-1\right)^2=0\\\left(y-3\right)^2=0\\2\left(z+1\right)^2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=1\\y=3\\z=-1\end{cases}}\)
Vậy \(\hept{\begin{cases}x=1\\y=3\\z=-1\end{cases}}\)
9x2 + y2 + 2z2 - 18x + 4z - 6y + 20 = 0
⇔ ( 9x2 - 18x + 9 ) + ( y2 - 6y + 9 ) + ( 2z2 + 4z + 2 ) = 0
⇔ ( 3x - 3 )2 + ( y - 3 )2 + 2( z + 1 )2 = 0
⇔ \(\hept{\begin{cases}3x-3=0\\y-3=0\\z+1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=3\\z=-1\end{cases}}\)
\(9x^2+y^2+2z^2-18x+4z-6y+20=0\)
\(\Leftrightarrow\left(9x^2-18x+9\right)+\left(y^2-6y+9\right)+\left(2z^2+4z+2\right)=0\)
\(\Leftrightarrow9\left(x^2-2x+1\right)+\left(y-3\right)^2+2\left(z^2+2z+1\right)=0\)
\(\Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\)
Vì \(\left(x-1\right)^2\ge0\forall x\)\(\Rightarrow9\left(x-1\right)^2\ge0\forall x\)
\(\left(y-3\right)^2\ge0\forall y\)
\(\left(z+1\right)^2\ge0\forall z\)\(\Rightarrow2\left(z+1\right)^2\ge0\forall z\)
\(\Rightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z-1\right)^2\ge0\forall x,y,z\)
Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}x-1=0\\y-3=0\\z+1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=3\\z=-1\end{cases}}\)
Vậy \(x=1\); \(y=3\); \(z=-1\)
Ta có:\(9x^2+y^2+2z^2-18x+4z-6y+20=0\)
\(9\left(\right.\) \(x^2-2x+1)+\) \(\left(\right.\) \(\left(y\right)^2-6y+\left(3\right)^2\) \()+2\) \(\left(\right.\) \(\left(z\right)^2+2z+1)=0\) \(\)
\(\Leftrightarrow9\left(x-1\right)^2+\left(y+3\right)^2+\left(z+1\right)^2=0\)
Mà\(\left(x-1\right)^2\ge0\)
\(\left(y+3\right)^2\ge0\)
\(\left(z+1\right)^2\ge0\)
\(\Rightarrow\) Dấu bằng xảy ra khi \(x-1=0\Rightarrow x=1\)
\(y+3=0\Rightarrow y=-3\)
\(z+1=0\Rightarrow z=-1\)