Ответ:
(1;2), (2;1)
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..[/tex]
Решим второе уравнение:
[tex]3^x+\dfrac{3^3}{3^x}=12;\ 3^x=t > 0;\ t+\dfrac{27}{t}=12;\ t^2-12t+27=0;\[/tex]
[tex](t-3)(t-9)=0;\ \left [ {{t=3} \atop {t=9}} \right. ;\ \left [ {{3^x=3^1} \atop {3^x=3^2}} \right.;\ \left [ {{x=1} \atop {x=2}} \right. .[/tex]
Если x=1⇒y=3-1=2; если x=2⇒y=3-2=1.
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Answers & Comments
Ответ:
(1;2), (2;1)
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..[/tex]
Решим второе уравнение:
[tex]3^x+\dfrac{3^3}{3^x}=12;\ 3^x=t > 0;\ t+\dfrac{27}{t}=12;\ t^2-12t+27=0;\[/tex]
[tex](t-3)(t-9)=0;\ \left [ {{t=3} \atop {t=9}} \right. ;\ \left [ {{3^x=3^1} \atop {3^x=3^2}} \right.;\ \left [ {{x=1} \atop {x=2}} \right. .[/tex]
Если x=1⇒y=3-1=2; если x=2⇒y=3-2=1.