What is x/(x^{2-1)}>0 ?

The solution to x/(x^{2-1)}>0 is x 0

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x/(x^{2-1)}>0

\frac{x}{x ^{2 - 1}} > 0

x < 0 or x > 0

Interval Notation

(- \infty, 0) \cup (0, \infty)

Solution steps

\frac{x}{x ^{2 - 1}} > 0

Simplify

1 > 0

True for all x

Find singularity points

Find the zeros of the denominator

x^{2 - 1} : x = 0

Since the inequality is not defined for

x = 0

x < 0 or x > 0

Combine the intervals

True for all x and x < 0 or x > 0

Merge Overlapping Intervals

x < 0 or x > 0

\frac{3}{8} \cdot \frac{3}{4}

- x^{2} - 5 x + 1 = 0

2 ∣ x + 4 ∣ = - 20

(2 x - 4) \cdot 2 + 2 \cdot (- 5 x + 1) = 12 - 6 x

\frac{2}{5 - 3}