\[ \frac{5}{9}+\frac{2}{9} = \frac{5+2}{9} = \boxed{\dfrac{7}{9}} \]

\[ \frac{11}{4}-\frac{3}{4} = \frac{11-3}{4} = \frac{8}{4} = \boxed{2} \]

Calculons \(\dfrac{2}{3}+\dfrac{1}{4}\) :

Le dénominateur commun : \(\text{PPCM}(3;\,4)=12\). \[ \frac{2}{3}+\frac{1}{4} = \frac{2\times 4}{3\times 4}+\frac{1\times 3}{4\times 3} = \frac{8}{12}+\frac{3}{12} = \frac{8+3}{12} = \boxed{\dfrac{11}{12}} \]

Calculons \(\dfrac{5}{6}-\dfrac{1}{9}\) :

\(\text{PPCM}(6;\,9)=18\). \[ \frac{5}{6}-\frac{1}{9} = \frac{5\times 3}{6\times 3}-\frac{1\times 2}{9\times 2} = \frac{15}{18}-\frac{2}{18} = \frac{15-2}{18} = \boxed{\dfrac{13}{18}} \]

Calculons \(\dfrac{7}{10}+\dfrac{3}{4}-\dfrac{1}{5}\) :

\(\text{PPCM}(10;\,4;\,5)=20\). \[ \frac{7}{10}+\frac{3}{4}-\frac{1}{5} = \frac{7\times 2}{20}+\frac{3\times 5}{20}-\frac{1\times 4}{20} = \frac{14}{20}+\frac{15}{20}-\frac{4}{20} \] \[ = \frac{14+15-4}{20} = \frac{25}{20} = \frac{25\div 5}{20\div 5} = \boxed{\dfrac{5}{4}} \]

\[ \frac{3}{5}\times\frac{2}{7} = \frac{3\times 2}{5\times 7} = \boxed{\dfrac{6}{35}} \]

Calculons \(\dfrac{4}{9}\times\dfrac{3}{8}\) :

Méthode 1 : Calcul direct puis simplification : \[ \frac{4}{9}\times\frac{3}{8} = \frac{4\times 3}{9\times 8} = \frac{12}{72} = \frac{12\div 12}{72\div 12} = \boxed{\dfrac{1}{6}} \] Méthode 2 : Simplification croisée avant de multiplier : \[ \frac{\overset{1}{\cancel{4}}}{{\overset{3}{\cancel{9}}}}\times\frac{\overset{1}{\cancel{3}}}{\underset{2}{\cancel{8}}} = \frac{1\times 1}{3\times 2} = \boxed{\dfrac{1}{6}} \]

Calculons \(\dfrac{5}{3}\times 4\) :

On écrit \(4=\dfrac{4}{1}\) : \[ \frac{5}{3}\times 4 = \frac{5}{3}\times\frac{4}{1} = \frac{5\times 4}{3\times 1} = \boxed{\dfrac{20}{3}} \]

L'inverse de \(\;\dfrac{3}{7}\;\) est \(\;\dfrac{7}{3}\)	car \(\;\dfrac{3}{7}\times\dfrac{7}{3}=\dfrac{21}{21}=1\;\checkmark\)
L'inverse de \(\;5\;\) est \(\;\dfrac{1}{5}\)	car \(\;5\times\dfrac{1}{5}=\dfrac{5}{5}=1\;\checkmark\)
L'inverse de \(\;0{,}25=\dfrac{1}{4}\;\) est \(\;4\)	car \(\;0{,}25\times 4=1\;\checkmark\)

Calculons \(\dfrac{3}{4}\div\dfrac{2}{5}\) : \[ \frac{3}{4}\div\frac{2}{5} = \frac{3}{4}\times\frac{5}{2} = \frac{3\times 5}{4\times 2} = \boxed{\dfrac{15}{8}} \]

Calculons \(\dfrac{7}{10}\div 3\) :

On écrit \(3=\dfrac{3}{1}\), son inverse est \(\dfrac{1}{3}\) : \[ \frac{7}{10}\div 3 = \frac{7}{10}\times\frac{1}{3} = \frac{7\times 1}{10\times 3} = \boxed{\dfrac{7}{30}} \]

Calculons \(\dfrac{14}{15}\div\dfrac{7}{9}\) : \[ \frac{14}{15}\div\frac{7}{9} = \frac{14}{15}\times\frac{9}{7} = \frac{\overset{2}{\cancel{14}}\times\overset{3}{\cancel{9}}}{\underset{5}{\cancel{15}}\times\underset{1}{\cancel{7}}} = \frac{2\times 3}{5\times 1} = \boxed{\dfrac{6}{5}} \]

Calculons \(\;A = \dfrac{2}{3}+\dfrac{1}{4}\times\dfrac{3}{5}\;\) :

Étape 1 : On effectue d'abord la multiplication (prioritaire) : \[ \frac{1}{4}\times\frac{3}{5}=\frac{3}{20} \] Étape 2 : Puis l'addition avec \(\text{PPCM}(3;\,20)=60\) : \[ A = \frac{2}{3}+\frac{3}{20} = \frac{2\times 20}{60}+\frac{3\times 3}{60} = \frac{40}{60}+\frac{9}{60} = \frac{49}{60} \] \[ \boxed{A=\dfrac{49}{60}} \]

Calculons \(\;B = \left(\dfrac{3}{2}-\dfrac{1}{3}\right)\times\dfrac{6}{7}\;\) :

Étape 1 : Les parenthèses d'abord, avec \(\text{PPCM}(2;\,3)=6\) : \[ \frac{3}{2}-\frac{1}{3} = \frac{9}{6}-\frac{2}{6} = \frac{7}{6} \] Étape 2 : Puis la multiplication : \[ B = \frac{7}{6}\times\frac{6}{7} = \frac{7\times 6}{6\times 7} = \frac{42}{42} = \boxed{1} \]

Calculer et simplifier : \[ A=\frac{3}{4}+\frac{5}{6} \qquad;\qquad B=\frac{7}{8}-\frac{2}{3} \] \[ C=\frac{5}{12}\times\frac{9}{10} \qquad;\qquad D=\frac{8}{15}\div\frac{4}{5} \] \[ E=\frac{1}{2}+\frac{3}{4}\times\frac{2}{5} \qquad;\qquad F=\left(\frac{5}{3}-\frac{1}{2}\right)\div\frac{7}{6} \]

► Calcul de A :   \(\text{PPCM}(4;\,6)=12\) \[ A = \frac{3}{4}+\frac{5}{6} = \frac{3\times 3}{12}+\frac{5\times 2}{12} = \frac{9}{12}+\frac{10}{12} = \frac{19}{12} \] \(\text{PGCD}(19;\,12)=1\), donc \(\;\boxed{A=\dfrac{19}{12}}\;\) est irréductible.

► Calcul de B :   \(\text{PPCM}(8;\,3)=24\) \[ B = \frac{7}{8}-\frac{2}{3} = \frac{7\times 3}{24}-\frac{2\times 8}{24} = \frac{21}{24}-\frac{16}{24} = \frac{5}{24} \] \[ \boxed{B=\dfrac{5}{24}} \] ► Calcul de C :   Simplification croisée : \(\gcd(5,10)=5\) et \(\gcd(9,12)=3\) \[ C = \frac{5}{12}\times\frac{9}{10} = \frac{\overset{1}{\cancel{5}}}{\underset{4}{\cancel{12}}}\times\frac{\overset{3}{\cancel{9}}}{\underset{2}{\cancel{10}}} = \frac{1\times 3}{4\times 2} \] \[ \boxed{C=\dfrac{3}{8}} \] ► Calcul de D : \[ D = \frac{8}{15}\div\frac{4}{5} = \frac{8}{15}\times\frac{5}{4} = \frac{\overset{2}{\cancel{8}}}{\underset{3}{\cancel{15}}}\times\frac{\overset{1}{\cancel{5}}}{\underset{1}{\cancel{4}}} = \frac{2\times 1}{3\times 1} \] \[ \boxed{D=\dfrac{2}{3}} \] ► Calcul de E :   Priorité à la multiplication : \[ \frac{3}{4}\times\frac{2}{5}=\frac{6}{20}=\frac{3}{10} \] \[ E = \frac{1}{2}+\frac{3}{10} = \frac{5}{10}+\frac{3}{10} = \frac{8}{10} = \frac{8\div 2}{10\div 2} \] \[ \boxed{E=\dfrac{4}{5}} \] ► Calcul de F :   Priorité aux parenthèses : \(\text{PPCM}(3;\,2)=6\) \[ \frac{5}{3}-\frac{1}{2} = \frac{10}{6}-\frac{3}{6} = \frac{7}{6} \] \[ F = \frac{7}{6}\div\frac{7}{6} = \frac{7}{6}\times\frac{6}{7} = \frac{42}{42} \] \[ \boxed{F=1} \]