Bài làm:
Ta có: \(x^2+2x+y^2-6y+4z^2-4z+11=0\)
\(\Leftrightarrow\left(x^2+2x+1\right)+\left(y^2-6y+9\right)+\left(4z^2-4z+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)^2+\left(y-3\right)^2+\left(2z-1\right)^2=0\)
Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left(x+1\right)^2=0\\\left(y-3\right)^2=0\\\left(2z-1\right)^2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-1\\y=3\\x=\frac{1}{2}\end{cases}}\)
Xin lỗi mk nhầm đoạn cuối là: \(\Rightarrow\hept{\begin{cases}x=-1\\y=3\\z=\frac{1}{2}\end{cases}}\) nhé:)
<=>x2+2x+1+y2-6y+9+4z2-4z+1=0
<=>(x2+2x+1)+(y2-6y+9)+(4z2-4z+1)=0
<=>(x+1)2+(y-3)2+(2z+1)2=0 (1)
Từ (1) <=> (x+1)2=0 <=> x+1=0 <=>x=-1
(y-3)2=0 <=>y-3=0 <=>y=3
(2z+1)2=0 <=>2z+1=0 <=> z=-1/2
x2 + 2x + y2 - 6y + 4z2 - 4z + 11 = 0
<=> ( x2 + 2x + 1 ) + ( y2 - 6y + 9 ) + ( 4z2 - 4z + 1 ) = 0
<=> ( x + 1 )2 + ( y - 3 )2 + ( 2z - 1 )2 = 0
\(\hept{\begin{cases}\left(x+1\right)^2\ge0\forall x\\\left(y-3\right)^2\ge0\forall y\\\left(2z-1\right)^2\ge0\forall z\end{cases}}\Rightarrow\left(x+1\right)^2+\left(y-3\right)^2+\left(2z-1\right)^2\ge0\forall x,y,z\)
Đẳng thức xảy ra <=> \(\hept{\begin{cases}x+1=0\\y-3=0\\2z-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-1\\y=3\\z=\frac{1}{2}\end{cases}}\)
