Soit [tex]A = \frac{1}{x+1} - \frac{1}{x+3} [/tex] pour x différent de -1 et de -3 Montrer que [tex]A = \frac{2}{(x+1)(x+3)} [/tex] Calculer A pour x = [tex]

Bonsoir,

[tex]A=\dfrac{1}{x+1}-\dfrac{1}{x+3}\\\\A=\dfrac{x+3}{(x+1)(x+3)}-\dfrac{x+1}{(x+1)(x+3)}\\\\A=\dfrac{(x+3)-(x+1)}{(x+1)(x+3)}\\\\A=\dfrac{x+3-x-1}{(x+1)(x+3)}\\\\A=\dfrac{2}{(x+1)(x+3)}[/tex]

Si x = -5/2, alors

[tex]A(-\dfrac{5}{2})=\dfrac{1}{-\dfrac{5}{2}+1}-\dfrac{1}{-\dfrac{5}{2}+3}\\\\A(-\dfrac{5}{2})=\dfrac{1}{-\dfrac{5}{2}+\dfrac{2}{2}}-\dfrac{1}{-\dfrac{5}{2}+\dfrac{6}{2}}\\\\A(-\dfrac{5}{2})=\dfrac{1}{-\dfrac{3}{2}}-\dfrac{1}{\dfrac{1}{2}}\\\\A(-\dfrac{5}{2})=1\times(-\dfrac{2}{3})-1\times2\\\\A(-\dfrac{5}{2})=-\dfrac{2}{3}-\dfrac{6}{3}\\\\A(-\dfrac{5}{2})=-\dfrac{8}{3}[/tex]

 ou encore 


[tex]A(-\dfrac{5}{2})=\dfrac{2}{(-\dfrac{5}{2}+1)(-\dfrac{5}{2}+3)}\\\\A(-\dfrac{5}{2})=\dfrac{2}{(-\dfrac{5}{2}+\dfrac{2}{2})(-\dfrac{5}{2}+\dfrac{6}{2})}\\\\A(-\dfrac{5}{2})=\dfrac{2}{(-\dfrac{3}{2})\times(\dfrac{1}{2})}\\\\\\A(-\dfrac{5}{2})=\dfrac{2}{-\dfrac{3}{4}}\\\\\\A(-\dfrac{5}{2})=2\times(-\dfrac{4}{3})\\\\\\A(-\dfrac{5}{2})=-\dfrac{8}{3}[/tex]