[tex]\it ^{6)}1-\dfrac{^{2)}1}{\ \ 3}-\dfrac{1}{6}=\dfrac{6-2-1}{6}=\dfrac{\ \ 3^{(3}}{6}=\dfrac{1}{2}[/tex]
Sau:
[tex]\it \dfrac{^{2)}1}{\ \ 3}+\dfrac{1}{6}=\dfrac{2+1}{6}=\dfrac{\ \ 3^{(3}}{6}=\dfrac{1}{2} \ din\ rond\ (panselu\c{\it t}e\ +\ lalele)\\ \\ \\ 1-\dfrac{1}{2}=\dfrac{1}{2}\ din\ rond\ (narcise)[/tex]