A=(x²+1)²+|2y-7|

To solve the inequality A = (x²+1)² + |2y-7| < 0, we need to consider two cases: when |2y-7| is positive and when |2y-7| is negative.

Case 1: |2y-7| > 0
In this case, the inequality becomes (x²+1)² + (2y-7) < 0.
Since (x²+1)² is always non-negative, for the inequality to hold, (2y-7) must be negative.
Solving 2y-7 < 0, we get y < 7/2.

Case 2: |2y-7| < 0
In this case, the inequality becomes (x²+1)² - (2y-7) < 0.
Since (x²+1)² is always non-negative, for the inequality to hold, (2y-7) must be positive.
Solving 2y-7 > 0, we get y > 7/2.

Combining the two cases, we have y < 7/2 or y > 7/2 as the solution to the inequality.