Question

РЕШИТЕ СИСТЕМЫ УРАВНЕНИЙ!!!!!ПОЖАЛУЙСТА!!!

Answer 1

А) $\left \{ {{log_{4}x-log_{4}y=1} \atop {x-3y=16}} \right.$

$\left \{ {{log_{4}x-log_{4}y=1} \atop {x=16+3y}} \right.$

$\left \{ {{log_{4}(16+3y)-log_{4}y=1} \atop {x=16+3y}} \right.$

$\left \{ {{log_{4}( \frac{16+3y}{y})=log_{4}4} \atop {x=16+3y}} \right.$

$\left \{ {{\frac{16+3y}{y}=4} \atop {x=16+3y}} \right.$

$\left \{ {{16+3y=4y} \atop {x=16+3y}} \right.$

$\left \{ {{y=16} \atop {x=16+3y}} \right.$

$\left \{ {{y=16} \atop {x=16+3*16=64}} \right.$

ОДЗ: x>0, y>0

Ответ: (64;16)

б) $\left \{ {{log_{2}(x-y)=1} \atop {2^{x}*3^{y+1}=72}} \right.$

$\left \{ {{x-y=2} \atop {2^{x}*3^{y}*3}=72}} \right.$

$\left \{ {{x=2+y} \atop {2^{x}*3^{y}*3}=72}} \right.$

$\left \{ {{x=2+y} \atop {2^{2+y}*3^{y}*3}=72}} \right.$

$\left \{ {{x=2+y} \atop {4*2^{y}*3^{y}*3}=72}} \right.$

$\left \{ {{x=2+y} \atop {2^{y}*3^{y}=6}} \right.$

$\left \{ {{x=2+y} \atop {6^{y}=6}} \right.$

$\left \{ {{x=2+y} \atop {y=1}} \right.$

$\left \{ {{x=2+1=3} \atop {y=1}} \right.$

ОДЗ: x-y>0

Ответ: (3;1)