√(2х²+7) = х+2
ОДЗ: 2х²+7 ≥ 0, т.е. при любом х.
2х²+7 = х²+4х+4
х²-4х+3 = 0
х1 = 1, х2 = 3
log₂(2x-1) = 3
2³ = 2x-1
8 = 2x-1
2x = 9
x = 4,5
/x-y = 40
\√x+√y = 10
/(√x-√y)(√x+√y) = 40
\√x+√y = 10
/√x-√y = 40/10 = 4
\√x+√y = 10
/√x = 4+√y
\4+√y+√y = 10
/√x = 4+3 = 7
\√y = 3
/x = 49
\y = 9
√(x+3) < x+1 (ОДЗ: х ≥ -3)
x+3 < x²+2x+1
x²-х-2 < 0
(x-2)(x+1) < 0
x ∈ (-1;2)
2lg6-lgx > 3lg2
lg36 > lgx-lg8
lg36 > lgx/8
36 > x/8
x < 288