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

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

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

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

Suy ra \(M=2\)

Ta có : 4x^2+2y^2+2z^2-4xy+2yz-6y-10z+34=04x2+2y2+2z2−4xy+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⇔(4x2+y2+z2−4xy−4xz+2yz)+(y2−6y+9)+(z2−10z+25)=0

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

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

Suy ra M=2M=2

Lời giải:
Ta có:

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

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

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

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

Vì \((2x-y-z)^2; (y-3)^2; (z-5)^2\geq 0, \forall x,y,z\). Do đó để \((2x-y-z)^2+(y-3)^2+(z-5)^2=0\) thì:
\((2x-y-z)^2=(y-3)^2=(z-5)^2=0\)

\(\Rightarrow \left\{\begin{matrix} x=4\\ y=3\\ z=5\end{matrix}\right.\)

Khi đó:

\(S=(4-4)^{2018}+(3-4)^{2019}+(5-4)^{2020}=0+(-1)+1=0\)

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:
\((2x-y-z)^2=(y-3)^2=(z-5)^2=0\Rightarrow \left\{\begin{matrix} y=3\\ z=5\\ x=4\end{matrix}\right.\)

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

thôi ko cần làm nữa đâu

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

4x² + 2y² + 2z²-4xy - 4xz+2yz-6y-10z+34=0
<=>(-2x+y+z)²+(y-3)²+(z-5)²=0
<=>\(\left\{{}\begin{matrix}-2x+y+z=0\\y=3\\z=5\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}x=4\\y=3\\z=5\end{matrix}\right.\)
Vậy A=\(\left(4-4\right)^{22}+\left(3-4\right)^6+\left(5-4\right)^{2013}=0^{22}+\left(-1\right)^6+1^{2013}=0+1+1=2\)