Réponses

2014-02-18T21:23:28+01:00
Bonsoir

a)\ \dfrac{1}{x}+1=\dfrac{2}{x}\\\\Condition : x\neq0\\\\\dfrac{1}{x}-\dfrac{2}{x}=-1\\\\-\dfrac{1}{x}=-1\\\\\dfrac{1}{x}=1\\\\x=1\\\\S=\{1\}

b)\ (\dfrac{1}{1+x})^2=1\\\\Condition:x\neq-1\\\\\dfrac{1}{(1+x)^2}=1\\\\(1+x)^2=1\\\\(1+x)^2-1=0\\\\(1+x)^2-1^2=0\\\\\ [(1+x)-1][(1+x)+1]=0

(1+x-1)(1+x+1)=0\\\\x(x+2)=0\\\\x=0\ \ ou\ \ x+2=0\\\\x=0\ \ ou\ \ x=-2\\\\S=\{0;-2\}
Merci:) mais j ai oublié la premiere equation c est (1/x^2)+1=2/x
Alors voici la réponse.
1/x² + 1 = 2/x
Condition : x différent de 0
1/x² + x²/x² = 2x/x²
(1 + x²)/x² = 2x/x²
1 + x² = 2x
1 + x² - 2x = 0
(x - 1)² = 0
x-1 = 0
x = 1