2x2 + 2y2 + z2 + 2xy + 2xz + 2yz + 10x + 6y + 34 = 0  

(x2 + y2 + z2 + 2xy + 2xz + 2yz) + (x2 + 10x + 25) + (y2+ 6y + 9) = 0  

( x + y + z)2 + ( x + 5)2 + (y + 3)2 = 0

( x + y + z)2 = 0 ;

( x + 5)2 = 0 ;

(y + 3)2 = 0

vậy x = - 5 ; y = -3; z = 8 

Tìm x, y, z biết rằng: 2x2 + 2y2 + z2 + 2xy + 2xz + 2yz + 10x + 6y + 34 = 0

                                Giải

2x2 + 2y2 + z2 + 2xy + 2xz + 2yz + 10x + 6y + 34 = 0 

(x2 + y2 + z2 + 2xy + 2xz + 2yz) + (x2 + 10x + 25) + (y2+ 6y + 9) = 0

 ( x + y + z)2 + ( x + 5)2 + (y + 3)2 = 0 

( x + y + z)2 = 0 ; ( x + 5)2 = 0 ; (y + 3)2 = 0

x = - 5 ; y = -3; z = 8 

Phần trả lời của mình cũng giống như Arcobaleno vậy

\(2x^2+2y^2+z^2+2xy+2yz+2xz+10x+6y+34=0\)

\(\Leftrightarrow\left(x^2+y^2+z^2+2xy+2yz+2xz\right)+\left(x^2+10x+25\right)+\left(y^2+6y+9\right)=0\)

\(\Leftrightarrow\left(x+y+z\right)^2+\left(x+5\right)^2+\left(y+3\right)^2=0\)

\(\Rightarrow x=-5,y=-3,z=8\)

Ta có:

\(\left(x^2+2xy+y^2\right)+\left(y^2+2yz+z^2\right)+\left(z^2+2zx+x^2\right)+\left(x^2+10x+25\right)+\left(y^2+6y+9\right)+z^2=0\)\(\Leftrightarrow\left(x+y\right)^2+\left(y+z\right)^2+\left(z+x\right)^2+\left(x+5\right)^2+\left(y+3\right)^2+z^2=0\)

Không tồn tại x,y,z thỏa mãn đề bài

\(2x^2+2y^2+z^2+2xy+2xz+2yz+10x+6y+34=0\)

\(\Leftrightarrow\left(x^2+y^2+z^2+2xy+2xz+2yz\right)+\left(x^2+10x+25\right)+\left(y^2+6y+9\right)=0\)

\(\Leftrightarrow\left(x+y+z\right)^2+\left(x+5\right)^2+\left(y+3\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y+z\right)^2=0\\\left(x+5\right)^2=0\\\left(y+3\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}z=8\\x=-5\\y=-3\end{matrix}\right.\)

Vậy x = -5; y = -3; z = 8

Liên quan gì đến tứ giác

yiouoiyy

\(2x^2+2y^2+z^2+2xy+2xz+2yz+10x+6y+34=0\)

\(\Leftrightarrow\left(x^2+y^2+z^2+2xy+2yz+2zx\right)+\left(x^2+10x+25\right)+\left(y^2+6y+9\right)=0\)

\(\Leftrightarrow\left(x+y+z\right)^2+\left(x+5\right)^2+\left(y+3\right)^2=0\)

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 \(Min\left(A\right)=8\Leftrightarrow\hept{\begin{cases}x=0\\y=-\frac{1}{2}\end{cases}}\)

\(2x^2+2y^2+z^2+2xy+2xz+2yz+10x+6y+34=0\)

\(\Rightarrow\left(x^2+y^2+z^2+2xy+2xz+2yz\right)+\left(x^2+10x+25\right)+\left(y^2+6y+9\right)=0\)

\(\Rightarrow\left(x+y+z\right)^2+\left(x+5\right)^2+\left(y+3\right)^2=0\)

Vì \(\left\{{}\begin{matrix}\left(x+y+z\right)^2\ge0\\\left(x+5\right)^2\ge0\\\left(y+3\right)^3\ge0\end{matrix}\right.\)

\(\Rightarrow\left(x+y+z\right)^2+\left(x+5\right)^2+\left(y+3\right)^2\ge0\)

Mà \(\left(x+y+z\right)^2+\left(x+5\right)^2+\left(y+3\right)^2=0\)

\(\Rightarrow\left\{{}\begin{matrix}\left(x+y+z\right)^2=0\\\left(x+5\right)^2=0\\\left(y+3\right)^2=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x+y+z=0\\x+5=0\\y+3=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}z=-\left(x+y\right)\\x=-5\\y=-3\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}z=8\\x=-5\\y=-3\end{matrix}\right.\)