Сделаем замену:
+\left \{ {{a+b=5} \atop {a-b=-3}} \right. =>2a=2 \\ a=1 \\ 1+b=5=>b=4 \\ \left \{ {{(x-2)^2=1} \atop {(x-2y)^2=4}} \right. => \left \{ {{ \left \{ {{x-2=1=>x=3} \atop {x-2=-1=>x=1}} \right. } \atop { \left \{ {{x-2y=2} \atop {x-2y=-2}} \right. }} \right.=> \left \{ {{x=3: \left \{ {{3-2y=2=>2y=1=>y=0,5} \atop {3-2y=-2=>2y=5=>y=2,5}} \right. } \atop {x=1: \left \{ {{1-2y=2=>2y=-1=>y=-0,5} \atop {1-2y=-2=>2y=3=>y=1,5}} \right. }} \right. => \\ " alt=" \left \{ {{(x-2)^2=a} \atop {(x-2y)^2=b}} \right. => +\left \{ {{a+b=5} \atop {a-b=-3}} \right. =>2a=2 \\ a=1 \\ 1+b=5=>b=4 \\ \left \{ {{(x-2)^2=1} \atop {(x-2y)^2=4}} \right. => \left \{ {{ \left \{ {{x-2=1=>x=3} \atop {x-2=-1=>x=1}} \right. } \atop { \left \{ {{x-2y=2} \atop {x-2y=-2}} \right. }} \right.=> \left \{ {{x=3: \left \{ {{3-2y=2=>2y=1=>y=0,5} \atop {3-2y=-2=>2y=5=>y=2,5}} \right. } \atop {x=1: \left \{ {{1-2y=2=>2y=-1=>y=-0,5} \atop {1-2y=-2=>2y=3=>y=1,5}} \right. }} \right. => \\ " align="absmiddle" class="latex-formula">
Ответ:
(3;0,5)
(3;2,5)
(1;-0,5)
(1;1,5)
