Tìm x,y,z biết :
a, (x-z)^2 + (y-z)^2 + y^2+z^2 = 2xy-2yz+6z-9
b, x^2 + 3y^2 + z^2 + 2xy-2yz-2x+4y+10=0
Tìm x,y,z biết : x2+3y2+z2+2xy-2yz-2x+4y+10=0
x^2+2xy+y^2+y^2-2yz+z^2+y^2+4y+4+6-2x=0
(x+y)^2+(y-z)^2+(y+2)^2+2*(3-x)=0
y+2=0=>y=-2
y-z=0=>z=-2
x+y=0=>x=2
<=>(x2+2xy+y2)+(y2-2yz+z2)+(y2+6y+9)-(2x+2y)+1=0
<=>[(x+y)2-2(x+y)+1]+(y-z)2+(y+3)2=0
<=>(x+y-1)2+(y-z)2+(y+3)2=0
Vì \(\hept{\begin{cases}\left(x+y-1\right)^2\ge0\\\left(y-z\right)^2\ge0\\\left(y+3\right)^2\ge0\end{cases}\Rightarrow\left(x+y-1\right)^2+\left(y-z\right)^2+\left(y+3\right)^2\ge0}\)
\(\Rightarrow\hept{\begin{cases}x+y-1=0\\y-z=0\\y+3=0\end{cases}\Rightarrow\hept{\begin{cases}x+y=1\\y-z=0\\y=-3\end{cases}}\Rightarrow\hept{\begin{cases}x=4\\z=-3\\y=-3\end{cases}}}\)
Vậy x=4,y=z=-3
Tìm x,y thỏa mãn (x-z)2+(y-z)2+y2+z2=2xy-2yz+6z-9
tìm x,y,z biết
2x^2 + 2y^2 +z^2 + 2xy + 2xz + 2yz + 10x + 6y + 34=0
tìm gtnn
A= 2x^2 + 4y^2 +4xy + 2x + 4y +9
\(2x^2+2y^2+z^2+2xy+2xz+2yz+10x+6y+34=0\)
\(\Leftrightarrow\left(x^2+y^2+z^2+2xy+2yz+2zx\right)+\left(x^2+10x+25\right)+\left(y^2+6y+9\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)^2+\left(x+5\right)^2+\left(y+3\right)^2=0\)
Vì \(\hept{\begin{cases}\left(x+y+z\right)^2\ge0\\\left(x+5\right)^2\ge0\\\left(y+3\right)^2\ge0\end{cases}}\)\(\Rightarrow\left(x+y+z\right)^2+\left(x+5\right)^2+\left(y+3\right)^2\ge0\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x+y+z\right)^2=0\\\left(x+5\right)^2=0\\\left(y+3\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x+y+z=0\\x+5=0\\y+3=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x+y+z=0\\x=-5\\y=-3\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-5\\y=-3\\z=8\end{cases}}}\)
\(A=2x^2+4y^2+4xy+2x+4y+9=\left(x^2+4y^2+4xy+2x+4y+1\right)+x^2+8\)
\(=\left(x+2y+1\right)^2+x^2+8\ge8\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x+2y+1=0\\x=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\y=-\frac{1}{2}\end{cases}}}\)
Vậy \(Min\left(A\right)=8\Leftrightarrow\hept{\begin{cases}x=0\\y=-\frac{1}{2}\end{cases}}\)
tìm x,y,z biết
\(a,2x^2+2y^2+z^2+2xy+2yz+10x+6y+34=0\)
\(b,x^2+y^2+z^2+2x-4y-6z+14=0\)
,\(c,2x^2+y^2-6x-4y+2xy+5=0\)
giúp mk với mn lm nhanh dùm mk đc hông sáng mai mk cần gấp
mk sẽ tick thật nhiều
b, x2 +y2+z2 +2x-4y-6z+14=0
<=> (x2+2x+1)+(y2-4y+4)+(z2-6z+9)=0
<=> (x+1)2+(y-2)2+(z-3)2=0
=>(x+1)2=(y-2)2=(z-3)2=0
=>x+1=y-2=z-3=0
=> x=-1; y=2; z=3
c, 2x2+y2-6x-4y+2xy+5=0
<=> (x2+y2+4+2xy-4x-4y)+(x2-2x+1)=0
<=> (x+y-2)2+(x-1)2=0
=> (x+y-2)2=(x-1)2=0
=>x+y-2=x-1=0
=>x=1; y=1
tìm x,y biết:
1) 5x2 + 3y2 + z2 - 4z + 6xy + 4z + 6 = 0
2) 2x2 + 2y2 + z2 + 2xy + 2xz + 2x + 4y + 5 = 0
3) 2x2 + 2y2 + z2 + 2xy +2xz + 2yz + 10x + 6y + 34 = 0
Tìm x,y, z biết:
2x2+2y2+z2+2xy+2xz+2yz+2x+4y+5=0
2x2 + 2y2 + z2 + 2xy + 2xz + 2yz + 2x + 4y + 5 = 0
<=> (x2 + y2 + z2 + 2xy + 2yz + 2xz) + (x2 + 2x + 1) + (y2 + 4y + 4) = 0
<=> (x + y + z)2 + (x + 1)2 + (y + 2)2 = 0
\(\Leftrightarrow\left\{{}\begin{matrix}x+y+z=0\\x+1=0\\y+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=-2\\z=3\end{matrix}\right.\)
Tìm x, y, z thỏa mãn đẳng thức sau: (x - z)2 + (y - z)2 + y2 + z2 = 2xy - 2yz + 6z + 9
gpt :
a.2x^2+y^2+z^2=xy+yz+zx
b.2x^2+2y^2+z^2+2xy+2yz+2zx+2x+4y+5=0
c,x^6-2x^3+x^2-2x+2=0
hình như em ghi sai đề rồi em nhé vì câu a không cũng 1 dạng sẽ không đưa về hằng đẳng thức được!
Tìm x,y,z thỏa mãn đẳng thức sau :
(x-z)2 + (y-z)2 +y2 +z2 = 2xy -2yz + 6z - 9