X^2-3>0, -x^2+2>0, -x^2-9>0, x^2-27<0, x^2-8>0

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вопрос

To solve each inequality, we need to find the values of x that make each statement true:

1) x^2 - 3 > 0
This inequality means that x^2 is greater than 3. So, x can be any real number greater than the square root of 3. The solution can be written as x ∈ (-∞, -√3) U (√3, ∞).

2) -x^2 + 2 > 0
This inequality means that -x^2 is greater than -2, which can be rewritten as x^2 < 2. So, x can be any real number that is less than the square root of 2. The solution can be written as x ∈ (-√2, √2).

3) -x^2 - 9 > 0
This inequality means that -x^2 is greater than 9, which can be rewritten as x^2 < -9. Since x^2 cannot be negative, there are no real solutions for this inequality.

4) x^2 - 27 < 0
This inequality means that x^2 is less than 27. So, x can be any real number between -√27 and √27. The solution can be written as x ∈ (-√27, √27).

5) x^2 - 8 > 0
This inequality means that x^2 is greater than 8. So, x can be any real number greater than the square root of 8. The solution can be written as x ∈ (-∞, -√8) U (√8, ∞).