Réponses

Meilleure réponse !
2014-05-15T18:42:42+02:00
1/a -1/b
= b/ab - a/ab
= (b-a)/ab

a/(a-b) + b/(a+b)
= a(a+b)/(a-b)(a+b) + b(a-b)/(a-b)(a+b)
= [a(a+b)+b(a-b)]/(a+b)(a-b)
= [a²+ab+ab-b²]/(a+b)(a-b)
= (a²+2ab-b²)/(a+b)(a-b)

2/3a + 5/6b
= 12b/(3ax6b) + 15a/(6bx3a)
= 4b/6ab + 5a/6ab
= (4b+5a)/6ab
2014-05-15T18:43:53+02:00
Bonsoir,

Tu peux utiliser les règles d'addition des fractions.

a)
\frac 1a - \frac 1b = \frac{1\times b - 1\times a}{ab} = \frac{b-a}{ab}

b)
\frac{a}{a-b} + \frac{b}{a+b} = \frac{a\left(a+b\right) + b\left(a-b\right)}{\left(a+b\right)\left(a-b\right)} = \frac{a^2+ab + ab-b^2}{\left(a+b\right)\left(a-b\right)} = \frac{a^2+2ab-b^2}{\left(a+b\right)\left(a-b\right)}

c)
\frac{2}{3a}+ \frac{5}{6b} =\frac{2\times 6b +5\times 3a}{3a\times 6b} = \frac{15a+12b}{18ab}=\frac{3\left(5a+4b\right)}{3\times 6ab} = \frac{5a+4b}{6ab}

Si tu as des questions, n'hésite pas ! =)