Question

КАКАЯ ПАР ЧИСЕЛ РЕШЕНИЕМ

Answer 1

$\displaystyle\bf\\\left \{ {{t^{2}+a^{2} =1 } \atop {a-2t=1}} \right.\\\\1)\\\\t=1 \ ; \ a=1\\\\\\\left \{ {{1^{2} +1^{2} =1} \atop {1-2\cdot1=1}} \right. \\\\\\ \left \{ {{2\neq 1} \atop {-1\neq 1}} \right. \\\\\\2)\\\\t=1 \ ; a=0\\\\\\\left \{ {{1^{2} +0^{2} =1} \atop {0-2\cdot 1=1}} \right. \\\\\\\left \{ {{1=1} \atop {-2\neq 1}} \right.$

$\displaystyle\bf\\3)\\\\t=-1 \ ; \ a=-1\\\\\\\left \{ {{(-1)^{2} +(-1)^{2} =1} \atop {-1-2\cdot(-1)=1}} \right. \\\\\\\left \{ {{2\neq 1} \atop {1= 1}} \right. \\\\\\4)\\\\t=0 \ ; \ a=1\\\\\\\left \{ {{0^{2} +1^{2} =1} \atop {1-2\cdot0=1}} \right. \\\\\\\left \{ {{1= 1} \atop {1=1}} \right.\\\\\\Otvet: \ t=0 \ ; \ a=1$