<=>(x²+y²+z²+2xy+2yz+2xz)+(x²+2x+1)+(y²+4y+4)=0

<=>(x+y+z)²+(x+1)²+(y+2)²=0

Mà \(\hept{\begin{cases}\left(x+y+z\right)^2\ge0\\\left(x+1\right)^2\ge0\\\left(y+2\right)^2\ge0\end{cases}\Rightarrow\left(x+y+z\right)^2+\left(x+1\right)^2+\left(y+2\right)^2\ge0}\)

=>\(\hept{\begin{cases}x+y+z=0\\x+1=0\\y+2=0\end{cases}\Rightarrow\hept{\begin{cases}z=3\\x=-1\\y=-2\end{cases}}}\)

\(2x^2+2y^2+z^2+2xy+2yz+2zx+2x+4y+5\)

\(=\left(x^2+y^2+z^2+2xy+2yz+2zx\right)+\left(x^2+2x+1\right)+\left(y^2+4y+4\right)\)

\(=\left(x+y+z\right)^2+\left(x+1\right)^2+\left(y+2\right)^2=0\)

Mà: \(\hept{\begin{cases}\left(x+y+z\right)^2\ge0\\\left(x+1\right)^2\ge0\\\left(y+2\right)^2\ge0\end{cases}}\Rightarrow\hept{\begin{cases}\left(x+y+z\right)^2=0\\\left(x+1\right)^2=0\\\left(y+2\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x+y+z=0\\x+1=0\\y+2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x+y+z=0\\x=-1\\y=-2\end{cases}}\Leftrightarrow\hept{\begin{cases}z=3\\x=-1\\y=-2\end{cases}}\)

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 \(\Leftrightarrow\hept{\begin{cases}\left(x+y+z\right)^2=0\\\left(x+5\right)^2=0\\\left(y+3\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x+y+z=0\\x+5=0\\y+3=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x+y+z=0\\x=-5\\y=-3\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-5\\y=-3\\z=8\end{cases}}}\)

g. G(x)=2x²+2y²+z²+2xy-2xz-2yz-2x-4y

           = [x²+2x(y-z)+(y²-2yz+z²)]+(x²-2x+1)+(y²-4y+4)-5

          = (x+y-z)²+(x-1)²+(y-2)²-5

Vì (x+y-z)²≥0∀x,y,z

     (x-1)²≥0∀x

      (y-2)²≥0∀y

⇒ G  = (x+y-z)²+(x-1)²+(y-2)²-5 ≥ -5

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x+y-z=0\\x-1=0\\y-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}z=3\\x=1\\y=2\end{matrix}\right.\)

h,H(x)=x² + y²-xy-x+y+1

⇔ 2H=2x²+2y²-2xy-2x-2y+2

         = (x²-2xy+y²)+(x²-2x+1)+(y²-2y+1)

         = (x-y)²+(x-1)²+(y-1)²

Vì (x-y)²≥0 ∀x,y

    (x-1)²≥0 ∀x

     (y-1)² ≥0 ∀y

⇒ 2H≥0 ⇒ H≥0

Dấu "=" xảy ra ⇔ x=y=1

cảm ơn bn

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 

\(B=\left(x^2+y^2+4+2xy-4x-4y\right)+\left(x^2+z^2+1+2xz-2x-2z\right)+\left(y^2-4y+4\right)+4\)

\(B=\left(x+y-2\right)^2+\left(x+z-1\right)^2+\left(y-2\right)^2+4\ge4\)

Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}x+y-2=0\\x+z-1=0\\y-2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=2\\z=1\end{matrix}\right.\)

\(G=2x^2+2y^2+z^2+2xy-2xz-2yz-2x-4y\)

\(=\left[x^2+2x\left(y-z\right)+\left(y-z\right)^2\right]+\left(x^2-2x+1\right)+\left(y^2-4y+4\right)-5\)

\(=\left(x+y-z\right)^2+\left(x-1\right)^2+\left(y-2\right)^2-5\ge-5\)

\(minG=-5\Leftrightarrow\) \(\left\{{}\begin{matrix}x+y-z=0\\x-1=0\\y-2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\\z=3\end{matrix}\right.\)

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

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

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

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