A)y=(x²-3x-4)(5x-x²)
D(y)∈(-∞;∞)
в)y=√(x²-3x-4)/√(5x-x²)
{x²-3x-4≥0 (1)
{5x-x²>0 (2)
(1) x1+x2=3 U x1*x2=-4⇒x1=-1 U x2=4
+ _ +
------------[-1]----------[4]--------------
x≤-1 U x≥4
(2) x(5-x)>0
x=0 x=5
_ + _
------------(0)---------(5)------------
0D(y)∈[4;5)