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Объяснение:

[tex]\left \{ {{5^{x+y}=125} \atop {3^x+3^y=12}} \right. \ \ \ \ \left \{ {{5^{x+y}=5^3} \atop {3^x+3^y=12}} \right. \ \ \ \ \ \left \{ {{x+y=3} \atop {3^x+3^y=12}} \right. \ \ \ \ \ \left \{ {{y=3-x} \atop {3^x+3^{3-x}=12}} \right. \ \ \ \ \ \left \{ {{y=3-x} \atop {3^x+\frac{3^3}{3^x}=12 }} \right. \\[/tex]

[tex]\left \{ {{y=3-x} \atop {3^{2x}+27=12*3^x}} \right. \ \ \ \ \ \left \{ {{y=3-x} \atop {3^{2x}-12*3^x+27=0}} \right..[/tex]

[tex]3^{2x}-12*3^x+27=0.[/tex]

Пусть 3ˣ=t>0       ⇒

[tex]t^2-12t+27=0\\D=36\ \ \ \ \sqrt{D}=6\\ t_1=3^x=3\\x_1=1.\\y=3-1=2.\\t-2=3^x=9=3^2\\x_2=2.\\y=3-2=1.[/tex]

Ответ: (1;2),  (2;1).