What is (x^2-81)/(x^4-81)<= 0 ?

The solution to (x^2-81)/(x^4-81)<= 0 is -9<= x<-3\lor 3<x<= 9

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(x^2-81)/(x^4-81)<= 0

\frac{x ^{2} - 81}{x ^{4} - 81} \leq 0

- 9 \leq x < - 3 or 3 < x \leq 9

Interval Notation

[- 9, - 3) \cup (3, 9]

Solution steps

\frac{x ^{2} - 81}{x ^{4} - 81} \leq 0

Factor

\frac{x ^{2} - 81}{x ^{4} - 81} : \frac{( x + 9 ) ( x - 9 )}{( x ^{2} + 9 ) ( x + 3 ) ( x - 3 )}

\frac{( x + 9 ) ( x - 9 )}{( x ^{2} + 9 ) ( x + 3 ) ( x - 3 )} \leq 0

Identify the intervals

- 9 \leq x < - 3 or 3 < x \leq 9

\frac{1}{3} k^{3} \frac{1}{n ^{7}}

- \frac{3}{10} x - 7 < \frac{1}{2}

- 5 x + 8 = - 7

\frac{x ^{- 3}}{x ^{2}}

3 x y - 4 x - 12 y + 16

What is (x^2-81)/(x^4-81)<= 0 ?
The solution to (x^2-81)/(x^4-81)<= 0 is -9<= x<-3\lor 3<x<= 9