\(\left|x-2,2\right|+\left|y+2,6\right|=0\)
\(\Rightarrow\left|x-2,2\right|=0\) và \(\left|y+2,6\right|=0\)
+) \(x-2,2=0\Rightarrow x=2,2\)
Vậy \(x=2,2\)
Ta có: Ix-2,2I \(_{\ge}\)0
Iy+2,6I \(\ge\)0
=> Ix-2,2I + Iy+2,6I \(\ge\)0
mà Ix-2,2I + Iy+2,6I = 0
nên Ix-2,2I = 0; Iy+2,6I = 0
=> x = 2,2 .