Réponse :
D=1
Explications étape par étape :
[tex]\frac{4d^{2}-(d-3)^{2} }{9(d^{2}-1) } -\frac{d^{2}-9}{(2d+3)^{2}-d^{2} } +\frac{(2d-3)^{2}-d^{2} }{4d^{2}-(d+3)^{2} } \\\frac{4d^{2}-(d-3)^{2} }{9(d-1)(d+1) } -\frac{(d-3)(d+3)}{(2d+3-d)(2d+3+d) } +\frac{(2d-3-d)(2d-3+d) }{(2d-d-3)(2d+d+3) }\\\frac{4d^{2}-(d-3)^{2} }{9(d-1)(d+1) } -\frac{(d-3)(d+3)}{3(d+3)(d+1) } +\frac{3(d-3)(d-1) }{3(d-3)(d+1) }\\\frac{4d^{2}-(d-3)^{2} }{9(d-1)(d+1) } -\frac{(d-3)}{3(d+1) } +\frac{(d-1) }{(d+1) }\\[/tex]
[tex]\frac{4d^{2}-(d-3)^{2} }{9(d-1)(d+1) } -\frac{3(d-3)(d-1)}{9(d-1)(d+1)} +\frac{9(d-1)^{2} }{9(d-1)(d+1)}\\\frac{4d^{2}-d^{2}+6d-9-3(d^{2}-d-3d+3)+9d^{2}-18d+9}{9(d-1)(d+1)} \\\frac{9d^{2}-9 }{9(d-1)(d+1)}=\frac{9(d-1)(d+1)}{9(d-1)(d+1)} =1 \\[/tex]
