Ответ:

a) (-1; 1)

б) (5; 15)

Объяснение:

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

б) [tex]\left \{ {{y+x^2=8x} \atop {log(3)(y)-log(3)(x)=1}} \right. \\\\y+x^2=8x\\y=8x-x^{2}\\\\log(3)(y)-log(3)(x)=1\\log(3)(8x-x^{2})-log(3)(x)=log(3)(3)\\log(3)(\frac{x(8-x)}{x} )=log(3)(3)\\8-x=3\\x=5\\\\\left \{ {{y=8x-x^{2}} \atop {x=5}} \right. \\\left \{ {{y=8*5-25=15} \atop {x=5}} \right.[/tex]