Question

нужно срочно решить систему уравнений
{log2(x+y)+2log4(x-y)=5}
{3^1+2log3(x-y)=48}
11 класс

Answer 1

log₂(x+y)+2*log₄(x-y)=5    ОДЗ: х+у>0     x-y>0   
3^(1+2*log₃(x-y))=48

log₂(x+y)+2*(1/2)*log₂(x-y)=5
3^(log₃27+log₃(x-y)²)=48

log₂(x+y)+log₂(x-y)=5
3^log₃(27*(x-y)²)=48

log₂((x+y)(x-y))=5
27*(x-y)²=48

log₂((x-y)(x+y))=5
(x-y)²=48/27=16/9

(x-y)(x+y)=2⁵=32
x-y=+/-4/3

(4/3)(x+y)=32   
x+y=24
x-y=4/3
2x=24⁴/₃
x₁=12²/₃   y₁=11¹/₃

(-4/3)(x+y)=32
x+y=-24
x-y=-4/3
2x==-24⁴/₃
x₂=-12²/₃ ∉ОДЗ    y=-11¹/₃ ∉ОДЗ.
Ответ: x₁=12²/₃   y₁=11¹/₃.