\(\left|x-\frac{2}{3}\right|+\left|y+\frac{5}{9}\right|=0\)
Vì \(\left|x-\frac{2}{3}\right|\ge0\)và \(\left|y+\frac{5}{9}\right|\ge0\)nên \(\left|x-\frac{2}{3}\right|+\left|y+\frac{5}{9}\right|\ge0\)
(Dấu "="\(\Leftrightarrow\)\(\left|x-\frac{2}{3}\right|=0\)và \(\left|y+\frac{5}{9}\right|=0\))
\(\Leftrightarrow\hept{\begin{cases}x=\frac{2}{3}\\y=\frac{-5}{9}\end{cases}}\)
vì \(\left|x-\frac{2}{3}\right|>0\)hoặc =0 ;\(\left|y+\frac{5}{9}\right|>0\)hoặc =o
mà\(\left|x-\frac{2}{3}\right|+\left|y+\frac{5}{9}\right|=0\)
nên |x-2/3| =0 và |y+5/9|=0
\(\Rightarrow\hept{\begin{cases}x-\frac{2}{3}=0\\y+\frac{5}{9}=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{2}{3}\\y=\frac{-5}{9}\end{cases}}}\)
Ta có : \(\left|x-\frac{2}{3}\right|\ge0\forall x\)
\(\left|y+\frac{5}{9}\right|\ge0\forall y\)
\(\Leftrightarrow\left|x-\frac{2}{3}\right|+\left|y+\frac{5}{9}\right|\ge0\forall x,y\)
Dấu " = " xảy ra khi : \(\hept{\begin{cases}x-\frac{2}{3}=0\\y+\frac{5}{9}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{2}{3}\\y=-\frac{5}{9}\end{cases}}\)
Vậy : ...