Réponse :
Bonsoir,
Explications étape par étape :
[tex]\left\{\begin{array}{ccc}cx+y&=&1\\x+cy&=&1\\\end {array} \right.\\\\\\[/tex]
1)
[tex]Si \ c=0\ alors\\\\\left\{\begin{array}{ccc}y&=&1\\x&=&1\\\end {array} \right.\\Une\ seule\ solution\ (1,1)\\[/tex]
2)
[tex]Si\ c\neq 0\ alors \\\\\left\{\begin{array}{ccc}cx+y&=&1\\cx+c^2y&=&c\\\end {array} \right.\\\\\\\left\{\begin{array}{ccc}y(c^2-1)&=&c-1\\cx+y&=&1\\\end {array} \right.\\\\\\\left\{\begin{array}{ccc}y(c-1)(c+1)&=&c-1\\cx+y&=&1\\\end {array} \right.\\\\[/tex]
a)
[tex]si\ c= 1\ alors\\\\\left\{\begin{array}{ccc}0&=&0\\x&=&1-y\\\end {array} \right.\\\\Une\ infinit\' e\ de\ solutions\\\\[/tex]
b)
[tex]si\ c=-1\ alors\\\\\left\{\begin{array}{ccc}y*0&=&1 \\\\x&=&1-cy\\\end {array} \right.\\\\\\\left\{\begin{array}{ccc}0&=&1 \\\\x&=&1-cy\\\end {array} \right.\\\\Equation\ impossible\ pas\ de \ solution\\[/tex]
[tex]si\ c\neq -1\ alors \\\\\\\left\{\begin{array}{ccc}y&=&\dfrac{1}{c+1} \\\\x&=&\dfrac{1}{c+1}\\\end {array} \right.\\\\Une\ seule\ solution\\[/tex]
