Question

Помогите, пожалуйста, решить систему уравнений!!!

Answer 1

Ответ: (2;1).

Объяснение:

$\displaystyle\\\left \{ {{3^x-3^y=6} \atop {3^{x+y}+3^{y-1}=28}} \right. \ \ \ \ \ \ \left \{ {{3^x-3^y=6} \atop {3^x*3^y+\frac{3^y}{3}=28\ |*3 }} \right. \ \ \ \ \ \ \left \{ {{3^x-3^y=6} \atop {3*3^x*3^y+3^y=28}} \right..$

Пусть: $3^x=t > 0\ \ \ \ \ \ 3^y=v > 0\ \ \ \ \ \ \Rightarrow$

$\displaystyle\\\left \{ {t-v=6} \atop {3tv+v=84}} \right.\ \ \ \ \ \ \left \{ {{t=v+6} \atop {3v*(v+6)+v=84}} \ \ \ \ \ \ \left \{ {{t=v+6} \atop {3v^2+18v+v=84}} \right. \right. \\\\\\\left \{ {{t=v+6} \atop {3v^2+19v-84=0}} \right. \ \ \ \ \ \ \left \{ {{t=v+6} \atop {3v^2-9v+28v-84=0}} \right. \ \ \ \ \ \ \left \{ {{t=v+6} \atop {3v*(v-3)+28*(v-3)=0}} \right.$

$\displaystyle\\\left \{ {{t=v+6} \atop {(v-3)*(3v+28)=0}\right.\ \ \ \ \ \ \left \{ {{t=9} \atop {v_1=3\ \ \ \ v_2=-\frac{28}{3} \notin}} \right. \ \ \ \ \ \ \left \{ {{t=v+6} \atop {v=3}} \right.\ \ \ \ \ \ \left \{ {{t=9} \atop {v=3}} \right. .\\\\\\\left \{ {{3^x=3^2} \atop {3^y=3^1}} \right. \ \ \ \ \ \ \left \{ {{x=2} \atop {y=1}} \right. .$