Объяснение:

[tex]a)\ \left \{ {{8^{-2x+y}=512} \atop {0,25^{x-y}=16}} \right.\ \ \ \ \left \{ {{(2^3)^{y-2x}=2^9 \atop {(\frac{1}{4})^{x-y}=2^4 }} \right. \ \ \ \ \left \{ {{2^{6y-3x}=2^9} \atop {(2^2)^{y-x}=2^4}} \right. \ \ \ \ \left \{ {{2^{6y-3x}=2^9} \atop {2^{2y-2x}=2^4}} \right. \ \ \ \ \\[/tex]

[tex]\left \{ {{6y-3x=9\ |:3} \atop {2y-2x=4\ |:2}} \right. \ \ \ \ \left \{ {{2y-x=3} \atop {y-x=2}}. \right.[/tex]

Вычитаем из первого уравнения второе:

[tex]y=1.\\8^{-2x+1}=512\\8*8^{-2x}=512\ |:8\\(2^3)^{-2x}=64\\2^{-6x}=2^6\\-6x=6\ |:(-6)\\x=-1.[/tex]

Ответ: (-1;1).

[tex]b)\ ODZ:\ x > 0,\ \ \ \ y > 0.\\\left \{ {{y+x^2=8x} \atop {log_3y-log_3x=1}} \right.\ \ \ \ \left \{ {{y+x^2=8x} \atop {log_3\frac{y}{x} =log_33}} \right. \ \ \ \ \left \{ {{y+x^2=8x} \atop {\frac{y}{x} =3}} \right. \ \ \ \ \left \{ {{3x+x^2=8x} \atop {y=3x}} \right.\ \ \ \ \left \{ {{x^2-5x=0} \atop {y=3x}} \right.[/tex]

[tex]\left \{ {{x*(x-5)=0} \atop {y=3x}} \right. \ \ \ \ \left \{ {{x_1=0\notin\ \ \ \ x_2=5} \atop {y_2=3*5=15}} \right. .[/tex]

Ответ: (5;15).