Y = x + 1 / x-3 (x > 0), which is 1 / x

A: use the check function
y=x+1/x-3,x>0
>=2√(x*1/x)-3
=2-3
=-1
The range is [- 1, + ∞)

Y = (x Λ 2-x + 3) / x, X belongs to the range of (- ∞, 0)

y=(x∧2-x+3)/x
=x-1+3/x
=x+3/x-1
Because X0
be
-Y = - x-3 / x + 1 > = 2 √ (- x) * - 3 / x) + 1 = 2 √ 3 + 1 (note that if - x > 0 - 3 / x > 0, a + b > = 2 √ AB formula can be used)
Namely
y

Y = (x ^ 2-x + 3) / x, X ∈ (- ∞, 0), the detailed process of calculating the range

y=(x²-x+3)/x
=x+3/x-1
x

Y = x + 1 / x-3 (x > 0), which is 1 / x

Y = x + 1 / x-3 (x > 0), which is 1 / x

Y = (x Λ 2-x + 3) / x, X belongs to the range of (- ∞, 0)

Y = (x ^ 2-x + 3) / x, X ∈ (- ∞, 0), the detailed process of calculating the range

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