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 <=> \(\hept{\begin{cases}x-y=0\\y-2z=0\\z-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=y=2\\z=1\end{cases}}\)
a) \(9x^2-8xy+8y^2-28x+28=0\)
\(\Leftrightarrow\left(2x^2-8xy+8y^2\right)+\left(7x^2-28x+28\right)=0\)
\(\Leftrightarrow2\left(x^2-4xy+4y^2\right)+7\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow2\left(x-2y\right)^2+7\left(x-2\right)^2=0\)(1)
Vì \(\left(x-2y\right)^2\ge0,\left(x-2\right)\ge0\)nên (1) tương đương với \(\hept{\begin{cases}x-2y=0\\x-2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\y=1\end{cases}}}\)
b) \(x^2+2y^2+5z^2+1=2\left(xy+2yz+z\right)\)
\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(y^2-4yz+4z^2\right)+\left(z^2-2z+1\right)=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-2z\right)^2+\left(z-1\right)^2=0\)(2)
Vì \(\left(x-y\right)^2\ge0,\left(y-2z\right)^2\ge0,\left(z-1\right)^2\ge0\)
Nên (2) tương đương với \(\hept{\begin{cases}x-y=0\\y-2z=0\\z-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\y=2\\z=1\end{cases}}}\)
a,\(9x^2-8xy+8y^2-28x+28=0\)
\(\left(2x^2-8xy+8y^2\right)+\left(7x^2-28x+28\right)=0\)
\(2\left(x^2-4xy+4y^2\right)+7\left(x^2-4x+4\right)=0\)
\(2\left(x-2y\right)^2+7\left(x-2\right)^2=0\)
\(2\left(x-2y\right)^2+7\left(x-2\right)^2\ge0\forall x,y\)
Dấu"="xảy ra khi\(\orbr{\begin{cases}7\left(x-2\right)^2=0\\2\left(x-2y\right)^2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\y=1\end{cases}}}\)
b,\(x^2+2y^2+5z^2+1=2\left(xy+2yz+z\right)\)
\(x^2+2y^2+5z^2+1=2xy+4yz+2z\)
\(x^2+2y^2+5z^2+1-2xy-4yz-2z=0\)
\(\left(x^2-2xy+y^2\right)+\left(y^2-4yz+4z^2\right)+\left(z^2-2z+1\right)=0\)
\(\left(x-y\right)^2+\left(y-2z\right)^2+\left(z-1\right)^2=0\)
\(\left(x-y\right)^2+\left(y-2z\right)^2+\left(z-1\right)^2\ge0\forall x,y,z\)
Dấu"="xảy ra khi \(\hept{\begin{cases}\left(z-1\right)^2=0\\\left(y-2z\right)^2=0\\\left(x-y\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}z=1\\y=2\\x=2\end{cases}}}\)
