a)

pt <=>   \(\left(2x^2-8xy+8y^2\right)+\left(7x^2-28x+28\right)=0\)

<=>   \(2\left(x-2y\right)^2+7\left(x-2\right)^2=0\)

TA luôn có:   \(2\left(x-2y^2\right)+7\left(x-2\right)^2\ge0\forall x;y\) 

=> DẤU "=" XẢY RA <=>   \(\hept{\begin{cases}2\left(x-2y\right)^2=0\\7\left(x-2\right)^2=0\end{cases}}\)

<=>   \(\hept{\begin{cases}y=1\\x=2\end{cases}}\)

b)

pt <=>   \(x^2+2y^2+5z^2-2xy-4yz-2z+1=0\)

<=>   \(\left(x^2-2xy+y^2\right)+\left(y^2-4yz+4z^2\right)+\left(z^2-2z+1\right)=0\)

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

LẬP LUẬN TƯƠNG TỰ NHƯ CÂU a ta cũng được:

DẤU "=" XẢY RA <=>   \(\left(x-y\right)^2=\left(y-2z\right)^2=\left(z-1\right)^2=0\)

=>   \(x=y=2;z=1\)

a) 9x² - 8xy + 8y² - 28x + 28 = 0

<=> ( 2x² - 8xy + 8y² ) + ( 7x² - 28x + 28 ) = 0

<=> 2( x² - 4xy + 4y² ) + 7( x² - 4x + 4 ) = 0

<=> 2( x - 2y )² + 7( x - 2 )² = 0

\(\hept{\begin{cases}2\left(x-2y\right)^2\ge0\forall x,y\\7\left(x-2\right)^2\ge0\forall x\end{cases}}\Rightarrow2\left(x-2y\right)^2+7\left(x-2\right)^2\ge0\forall x,y\)

Đẳng thức xảy ra <=> \(\hept{\begin{cases}x-2y=0\\x-2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=1\end{cases}}\)

b) x² + 2y² + 5z² + 1 = 2( xy + 2yz + z )

<=> x² + 2y² + 5z² + 1 = 2xy + 4yz + 2z

<=> x² + 2y² + 5z² + 1 - 2xy - 4yz - 2z = 0

<=> ( x² - 2xy + y² ) + ( y² - 4yz + 4z² ) + ( z² - 2z + 1 ) = 0

<=> ( x - y )² + ( y - 2z )² + ( z - 1 )² = 0

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

Đẳng thức xảy ra <=> \(\hept{\begin{cases}x-y=0\\y-2z=0\\z-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=y=2\\z=1\end{cases}}\)