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đề bài sai nhé, 6x phảy là 6y
\(\Leftrightarrow\left(4x^2+y^2+z^2-4xy-4xz+2yz\right)+\left(y^2-6y+9\right)+\left(z^2-10z+25\right)=0\)
\(\Leftrightarrow\left(-2x+y+z\right)^2+\left(y-3\right)^2+\left(z-5\right)^2=0\)
Vì \(\left(-2x+y+z\right)^2\ge0\)
\(\left(y-3\right)^2\ge0\)
\(\left(z-5\right)^2\ge0\)
\(\Rightarrow\left(-2x+y+z\right)^2+\left(y-3\right)^2+\left(z-5\right)^2\ge0\)
Dấu "=" xảy ra \(\Leftrightarrow y=3;z=5;x=4\)
\(\left(x-4\right)^{2015}+\left(y-4\right)^{2015}+\left(z-4\right)^{2015}=\left(4-4\right)^{2015}+\left(3-4\right)^{2015}+\left(5-4\right)^{2015}=0\)

Lời giải:
\(4x^2+2y^2+2z^2-4xy-4xz+2yz-6y-10z+34=0\)

\(\Leftrightarrow (4x^2-4xy+y^2)+y^2+2z^2-2z(2x-y)-6y-10z+34=0\)

\(\Leftrightarrow (2x-y)^2-2z(2x-y)+z^2+y^2+z^2-6y-10z+34=0\)

\(\Leftrightarrow (2x-y-z)^2+(y^2-6y+9)+(z^2-10z+25)=0\)

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

Do \((2x-y-z)^2; (y-3)^2; (z-5)^2\geq 0, \forall x,y,z\), nên để tổng của chúng bẳng $0$ thì:
\((2x-y-z)^2=(y-3)^2=(z-5)^2=0\Rightarrow \left\{\begin{matrix} y=3\\ z=5\\ x=4\end{matrix}\right.\)

\(\Rightarrow S=(x-4)^{2014}+(y-4)^{2015}+(z-4)^{2016}=0+(-1)^{2015}+1^{2016}=-1+1=0\)

Lời giải:
$4x^2+2y^2+2z^2-4xy-4xz+2yz-6y-10z+34=0$

$(4x^2+y^2+z^2-4xy-4xz+2yz)+y^2+z^2-6y-10z+34=0$

$(2x-y-z)^2+(y^2-6y+9)+(z^2-10z+25)=0$
$(2x-y-z)^2+(y-3)^2+(z-5)^2=0$

Vì $(2x-y-z)^2\geq 0; (y-3)^2\geq 0; (z-5)^2\geq 0$ với mọi $x,y,z$

Do đó để tổng của chúng bằng $0$ thì bản thân mỗi số đó bằng $0$

$\Rightarrow 2x-y-z=y-3=z-5=0$

$\Rightarrow y=3; z=5; x=4$

Khi đó:
$P=0^{2023}+(-1)^{2025}+(5-4)^{2027}=0$

Ta có : \(4x^2+2y^2+2z^2-4xy-4xz+2yz-6y-10z+34=0\)

    \(\Leftrightarrow\left(4x^2+y^2+z^2-4xy-4xz+2yz\right)+\left(y^2-6y+9\right)+\left(z^2-10z+25\right)=0\)

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

Do \(\hept{\begin{cases}\left(2x-y-z\right)^2\ge0\\\left(y-3\right)^2\ge0\\\left(z-5\right)^2\ge0\end{cases}\Rightarrow VT\ge0}\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}2x-y-z=0\\y-3=0\\z-5=0\end{cases}\Leftrightarrow\hept{\begin{cases}2x=y+z\\y=3\\z=5\end{cases}\Leftrightarrow}\hept{\begin{cases}x=4\\y=3\\z=5\end{cases}}}\)

Khi đó \(P=\left(4-4\right)^{2018}+\left(3-4\right)^{2018}+\left(5-4\right)^{2018}\)

               \(=0+\left(-1\right)^{2018}+1^{2018}\)

               \(=2\)

4x² + 2y² + 2z² - 4xy + 2yz - 4xz - 6y - 10z + 34 = 0

<=> [ ( 4x² - 4xy + y² ) - 4xz + 2yz + z² ] + ( y² - 6y + 9 ) + ( z² - 10z + 25 ) = 0

<=> [ ( 2x - y )² - 2( 2x - y )z + z² ] + ( y - 3 )² + ( z - 5 )² = 0

<=> ( 2x - y - z )² + ( y - 3 )² + ( z - 5 )² = 0

\(\hept{\begin{cases}\left(2x-y-z\right)^2\\\left(y-3\right)^2\\\left(z-5\right)^2\end{cases}}\ge0\forall x,y,z\Rightarrow\left(2x-y-z\right)+\left(y-3\right)^2+\left(z-5\right)^2\ge0\forall x,y,z\)

Đẳng thức xảy ra <=> \(\hept{\begin{cases}2x-y-z=0\\y-3=0\\z-5=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=4\\y=3\\z=5\end{cases}}\)

Thế vào S ta được :

S = ( x - 4 )²⁰²⁰ + ( y - 3 )²⁰²⁰ + ( z - 5 )²⁰²⁰

    = ( 4 - 4 )²⁰²⁰ + ( 3 - 3 )²⁰²⁰ + ( 5 - 5 )²⁰²⁰

    = 0 + 0 + 0

    = 0