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

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

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

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

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

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

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

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

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

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

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

Ta có:

D=2x2+3y2+4xy−8x−2y+18C=2x2+3y2+4xy−8x−2y+18

D=2(x2+2xy+y2)+y2−8x−2y+18C=2(x2+2xy+y2)+y2−8x−2y+18

D=2[(x+y)2−4(x+y)+4]+(y2+6y+9)+1C=2[(x+y)2−4(x+y)+4]+(y2+6y+9)+1

D=2(x+y−2)2+(y+3)2+1≥1C=2(x+y−2)2+(y+3)2+1≥1

Dấu "=" xảy ra ⇔x+y=2⇔x+y=2và y=−3y=−3

Hay x = 5 , y = -3

Đc chx bạn

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

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

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