Bonjour les matheux je cherche comment faire ces calcule si quelqun peut m'expliquer en détaillant svp ce serais treés gentils de votre part

Bonjour,

1)

On suppose a ≠ 0

[tex]\dfrac{?}{6a} +b=\dfrac{3a+5}{3a} \\\\\Longrightarrow\ \dfrac{?+6ab}{6a} =\dfrac{6a+10}{6a}\\\\\Longrightarrow\ ?+6ab =6a+10\\\\\boxed{?=6a+10-6ab}\\\\[/tex]

2)

On suppose q≠0

[tex]\dfrac{1}{10q} +\dfrac{1}{12q}+\dfrac{1}{6q}=\dfrac{?}{20q}\\\\\dfrac{1}{10q} +\dfrac{1}{12q}+\dfrac{2}{12q}=\dfrac{?}{20q}\\\\\dfrac{1}{10q} +\dfrac{3}{12q}=\dfrac{?}{20q}\\\\\dfrac{1}{10q} +\dfrac{1}{4q}=\dfrac{?}{20q}\\\\\dfrac{2}{20q} +\dfrac{5}{20q}=\dfrac{?}{20q}\\\\\dfrac{7}{20q}=\dfrac{?}{20q}\\\\\boxed{?=7}\\[/tex]

3)

On suppose r≠0

[tex]\dfrac{r}{10} +\dfrac{r}{12} =\dfrac{11r}{?} \\\\\dfrac{1}{10} +\dfrac{1}{12} =\dfrac{11}{?} \\\\\dfrac{6}{60} +\dfrac{5}{60} =\dfrac{11}{?} \\\\\dfrac{11}{60} =\dfrac{11}{?} \\\\\boxed{?=60}\\[/tex]

4)

On suppose n≠0 et p≠0

[tex]\dfrac{?}{6n} +\dfrac{10}{p} =\dfrac{2p+10n}{np} \\\\\dfrac{?*p}{6np} +\dfrac{60n}{6np} =\dfrac{12p+60n}{6np} \\\\?*p+60n =12p+60n \\\\?*p =12p\\\\\boxed{?=12}\\[/tex]