Réponse :
Explications étape par étape
[tex]A =\frac{1}{x} + \frac{2}{x+1} =\frac{x+1 +2x}{x(x+1)} = \frac{3x+1}{x(x+1)}\\\\B=\frac{1}{1-3x} - \frac{1}{1+3x} = \frac{1+3x - (1-3x)}{(1-3x)(1+3x)}=\frac{6x}{1-9x^{2}}\\\\C=\frac{2}{x+1} - \frac{(1-\frac{1}{x})}{1+\frac{1}{x}} = \frac{2(1+\frac{1}{x})-(x+1)(1-\frac{1}{x}) }{(x+1)(1+\frac{1}{x}) } \\\\=[2+ [2/x] -(x-1+1-[1/x])] / [x+1+1+1/x] \\\\=[2-x+3/x]/[x+2+1/x]\\\\=[-x^{2}+2x+3]/[x^{2}+2x+1][/tex]
