Réponse :
Bonjour,
Explications étape par étape :
[tex]\left\{\begin{array}{ccc}x^2-y^2-2x&=&-1\\xy+y&=&0\end {array} \right.\\\\\\\left\{\begin{array}{ccc}y(x+1)&=&0\\x^2-y^2-2x&=&-1\\\end {array} \right.\\\\\\\left\{\begin{array}{ccc}y&=&0\\x^2-2x&=&-1\\\end {array} \right. \ ou \ \left\{\begin{array}{ccc}x=&-1\\1-y^2+2&=&-1\\\end {array} \right.\\\\\\\left\{\begin{array}{ccc}y&=&0\\x&=&1\\\end {array} \right. \ ou \ \left\{\begin{array}{ccc}x=&-1\\y=2\ &ou& \ y=-2 \\\end {array} \right.\\[/tex]
[tex]\\(1,0)\ est\ solution\ car\ 1-0-2=-1\ et\ 1*0+0=0\\\\(-1,2)\ est\ solution\ car\ 1-4+2=-1\ et\ -1*2+2=0\\\\(-1,-2)\ est\ solution\ car\ 1-4+2=-1\ et\ -1*(-2)+(-2)=0\\[/tex]
