Réponse :
Bonsoir,
Je te laisse résoudre (S2).
[tex]\left\{\begin{array}{ccc}ax+y&=&2\\(a^2+1)x+2ay&=&1\end{array}\right.\\\\\Delta=\begin{array}{|cc|}a&1\\a^2+1&2a\end{array}=2a^2-a^2-1=a^2-1=(a-1)(a+1)\\\\\\\Delta_1=\begin{array}{|cc|}2&1\\1&2a\end{array}=4a-1\\\\\\\Delta_2=\begin{array}{|cc|}a&2\\a^2+1&1\end{array}=a-2a^2-2=-(2a^2-a+2)\\\\\\Si\ a=-1\ alors \\\left\{\begin{array}{ccc}-x+y&=&2\\2x-2y&=&1\end{array}\right.\\\\\Longrightarrow\ 0x+0y=5\ pas\ de\ solution\\[/tex]
[tex]Si\ a=1\ alors \\\left\{\begin{array}{ccc}x+y&=&2\\2x+2y&=&1\end{array}\right.\\\\\Longrightarrow\ 0x+0y=-3\ pas\ de\ solution\\\\Si\ a\neq -1\ et\ a\neq 1\ alors \\\\\boxed{\left\{\begin{array}{ccc}x&=&\dfrac{4a-1}{(a-1)(a+1)}\\\\y&=&\dfrac{2a^2-a+2}{(a-1)(a+1)}\end{array}\right.}[/tex]
