ta có:
(x - 1/2)2 \(\ge\) 0
(x + y)2 \(\ge\) 0
=> để (x - 1/2)2 + (x + y)2 = 0 thì :
(x - 1/2)2 = 0 và (x + y)2 = 0
(x - 1/2)2 = 0
=> x - 1/2 = 0 => x = 0 + 1/2 = 1/2
(x + y)2 = 0
=> x + y = 0 => 1/2 + y = 0 => y = 0 - 1/2 = -1/2
Vậy x = 1/2, y = -1/2
\(\left(x-\frac{1}{2}\right)^2+\left(x+y\right)^2=0\)
Ta có : \(\hept{\begin{cases}\left(x-\frac{1}{2}\right)^2\ge0\forall x\\\left(x+y\right)^2\ge0\forall x,y\end{cases}\Rightarrow}\left(x-\frac{1}{2}\right)^2+\left(x+y\right)^2\ge0\forall x,y\)
Dấu = xảy ra <=> \(\hept{\begin{cases}x-\frac{1}{2}=0\\x+y=0\end{cases}}\)
\(x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{2}\)
\(x+y=0\Leftrightarrow\frac{1}{2}+y=0\Leftrightarrow y=-\frac{1}{2}\)
\(\left(x-\frac{1}{2}\right)^2+\left(x+y\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}\left(x-\frac{1}{2}\right)^2\ge0\\\left(x+y\right)^2\ge0\end{cases}}\)
muốn có \(\left(x-\frac{1}{2}\right)^2+\left(x+y\right)^2=0\)
thì \(\orbr{\begin{cases}\left(x-\frac{1}{2}\right)^2=0\\\left(x+y\right)^2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x-\frac{1}{2}=0\\x+y=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\thay\frac{1}{2}+y=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\y=-\frac{1}{2}\end{cases}}\)
vậy x=1/2 và y= -1/2
ko biết????!!!!!!!..........
