Ta có:

\(2yz^2-2-4y^2z+5yz-A=0\)

\(\Leftrightarrow A=2yz^2-4y^2z+5yz-2\)

Vậy: ...

a) \(\Leftrightarrow4x^2+2y^2+4xy-20x-8y+26=0\)

\(\Leftrightarrow4x^2+4x\left(y-5\right)+\left(y-5\right)^2-\left(y-5\right)^2+2y^2-8y+26=0\)

\(\Leftrightarrow\left(2x+y-5\right)^2+y^2+2y+1=0\)

\(\Leftrightarrow\left(2x+y-5\right)^2+\left(y+1\right)^2=0\)

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

b) \(\Leftrightarrow\left(x^2-4x+4\right)+\left(y^2+6y+9\right)+\left(z^2-2z+1\right)=0\)

\(\Leftrightarrow\left(x-2\right)^2+\left(y+3\right)^2+\left(z-1\right)^2=0\)

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

c) \(\Leftrightarrow\left(x^2+y^2+z^2+2xy+2yz+2xz\right)+\left(x^2+2x+1\right)+\left(z^2-4z+4\right)=0\)

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

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

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

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

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

x=1;y=-1;z=2 nhé bn đấy là tìm mò còn lời giải để mình nghĩ cái ( hơi lâu đấy =((( )

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 \(\Leftrightarrow y=3;z=5;x=4\)
\(\left(x-4\right)^{2015}+\left(y-4\right)^{2015}+\left(z-4\right)^{2015}=\left(4-4\right)^{2015}+\left(3-4\right)^{2015}+\left(5-4\right)^{2015}=0\)