Bonjour Fokker974
2b) Compléter à l'aide d'une calculatrice.
[tex]\dfrac{5}{8}=\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{.}\\\\\dfrac{1}{.}=\dfrac{5}{8}-\dfrac{1}{3}-\dfrac{1}{6}\\\\\dfrac{1}{.}=\dfrac{1}{8}\\\\Donc\ \boxed{\dfrac{5}{8}=\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{8}}[/tex]
[tex]\dfrac{1}{12}+\dfrac{1}{.}+\dfrac{1}{68}=\dfrac{2}{17}\\\\\dfrac{1}{.}=\dfrac{2}{17}-\dfrac{1}{12}-\dfrac{1}{68}\\\\\dfrac{1}{.}=\dfrac{1}{51}\\\\Donc\ \ \boxed{\dfrac{1}{12}+\dfrac{1}{51}+\dfrac{1}{68}=\dfrac{2}{17}}[/tex]
[tex]\dfrac{4}{5}=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{.}\\\\\dfrac{1}{.}=\dfrac{4}{5}-\dfrac{1}{2}-\dfrac{1}{4}\\\\\dfrac{1}{.}=\dfrac{1}{20}\\\\Donc\ \ \boxed{\dfrac{4}{5}=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{20}}[/tex]
[tex]\dfrac{1}{.}+\dfrac{1}{3}+\dfrac{1}{9}=\dfrac{17}{18}\\\\\dfrac{1}{.}=\dfrac{17}{18}-\dfrac{1}{3}-\dfrac{1}{9}\\\\\dfrac{1}{.}=\dfrac{1}{2}\\\\Donc\ \ \boxed{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{9}=\dfrac{17}{18}}[/tex]
