It is known that the domain of F (x) = (MX-1) √ (MX ^ - 4mx + 5) is the value range of R for M f(x)=(mx-1)/√(mx^-4mx+5)

It is known that the domain of F (x) = (MX-1) √ (MX ^ - 4mx + 5) is the value range of R for M f(x)=(mx-1)/√(mx^-4mx+5)

When m = 0, f (x) = - 1 / √ 5 = - √ 5 / 5
The domain is r
When m ≠ 0, because √ (MX ^ 2-4mx + 5) is radical and denominator, it should be always greater than 0 when x ∈ R
So m > 0, △ = (- 4m) ^ 2-4 × 5 × m > 0
The solution is m ∈ (5 / 4, + ∞)
In conclusion, M = 0 or m ∈ (5 / 4, + ∞)