Bonsoir,
[tex]\dfrac{2}{x-1} - \dfrac{3}{x+1} = \dfrac{4}{x^2 -1}\\\\Valeurs\ interdites:x=1\ et\ x=-1\\\\\dfrac{2(x+1)}{(x-1)(x+1)} - \dfrac{3(x-1)}{(x+1)(x-1)} = \dfrac{4}{(x+1)(x-1)}\\\\\dfrac{2(x+1) -3(x-1)}{(x+1)(x-1)} = \dfrac{4}{(x+1)(x-1)}\\\\2(x+1)-3(x-1)=4\\\\2x+2-3x+3=4\\\\-x+5=4[/tex]
[tex]\\\\-x=4-5\\\\-x=-1\\\\x=1\\\\Valeur\ interdite\\\\Pas\ de\ solution\\\\S=\Phi[/tex]