Question

Помогите решить через ОДЗ

Answer 1

Пошаговое объяснение:

$a)\ x>0\ \ \ \ y>0.\\\left \{ {{log_8x+log_4y+3y=11} \atop {log_8x+log_4y+y=3}} \right. \ \ \ \ \left \{ {{log_8x+log_4y=11-3y} \atop {log_8x+log_4y=3-y}} \right.\ \ \ \ \Rightarrow\\11-3y=3-y \\2y=8\ |:2\\y=4.\ \ \ \ \Rightarrow\\log_8x+log_44+4=3\\log_8x+1=-1\\log_8x=-2\\x=8^{-2}=\frac{1}{64}.$

Ответ: (1/64;4).

$b)\\\left \{ {{log_3(5xy)-log_3(-5x)=2} \atop {x^2+5y+51=0}} \right. .$

ОДЗ: -5х>0 |:-5     x<0   5xy>0|:5     xy>0    y<0.

$\left \{ {{log_3\frac{5xy}{-5x} =2} \atop {x^2+5y+41=0}} \right. \ \ \ \ \left \{ {{log_3(-y)=2} \atop {x^2+5y+41=0}} \right. \ \ \ \ \left \{ {{-y=3^2\ |*(-1)} \atop {x^2+5y+41=0}} \right. \ \ \ \ \ \left \{ {{y=-9} \atop {x^2+5*(-9)+41=0}} \right.$

$\left \{ {{y=-9} \atop {x^2-45+41=0}} \right. \ \ \ \ \left \{ {{y=-9} \atop {x^2-4=0}} \right.\ \ \ \ \left \{ {{y=-9} \atop {x^2-2^2=0}} \right.\ \ \ \ \ \left \{ {{y=-9} \atop {(x+2)*(x-2)=0}} \right. \ \ \ \ \ \left \{ {{y=-9} \atop {x_1=-2\ \ \ \ x_2-2\notin}} \right.$

Ответ: (-2;-9).