Log₂(x+y)+2*log₄(x-y)=5 ОДЗ: x>y x>-y
3^(1+2*log₃(x-y)=48
log₂(x+y)+2*log₂²(x-y)=5
3*3^log₃(x-y)²=48
log₂(x+y)+2*(1/2)*log₂(x-y)=5
3*(x-y)²=48 |÷3
log₂(x+y)+log₂(x-y)=5
(x-y)²=16
1)
log₂((x+y)*(x-y))=5*log₂2
x-y=4
log₂(x²-y²)=log₂2⁵
y=x-4
x²-y²=32
y=x-4
x²-(x-4)²=32
x²-x²+8x-16=32
8x=48 |÷8
x=6 ⇒
y=6-4=2
2)
x²-y²=32
x-y=-4
x²-y²=32
y=x+4
x²-(x+4)²=32
x²-x²-8x-16=32
-8x=48 |÷(-8)
x=-6 ⇒
y=-2 ∉ ОДЗ
<span>Ответ: x=6 y=2.</span>
25x²-10x+1-4=0
(5x-1)²-4=0
(5x-1)²-2²=0
(5x-1-2)(5x-1+2)=0
(5x-3)(5x+1)=0
x1=3/5, x2= -1/5 или x1=0,6, x2= -0,2
----------------------------