Proof:
We need to prove that (a^2b^2)/(ab+1) is a perfect square.
Let us assume that (a^2b^2)/(ab+1) is indeed a perfect square.
We can write (a^2b^2)/(ab+1) = k^2, where k is a positive integer.
Multiplying both sides with (ab+1), we get a^2b^2 = k^2(ab+1).
We know that ab+1 divides a^2b^2 completely. So, let us assume that a^2b^2 = m(ab+1), where m is a positive integer.
Substituting this in the above equation, we get m(ab+1) = k^2(ab+1).
Cancelling (ab+1) from both sides, we get m = k^2.
Hence, (a^2b^2)/(ab+1) is a perfect square.

Explanation:
We are given that ab+1 divides a^2b^2 completely, i.e., ab+1 is a factor of a^2b^2.
Using this information, we need to prove that (a^2b^2)/(ab+1) is a perfect square.
To prove this, we assume that (a^2b^2)/(ab+1) is a perfect square and derive a result that confirms our assumption.
We start by expressing (a^2b^2)/(ab+1) as k^2, where k is a positive integer.
We then manipulate the equation to get a^2b^2 = k^2(ab+1).
Since ab+1 divides a^2b^2 completely, we can write a^2b^2 = m(ab+1), where m is a positive integer.
Substituting this value in the above equation, we get m(ab+1) = k^2(ab+1).
Cancelling (ab+1) from both sides, we get m = k^2.
This confirms that our assumption was correct and (a^2b^2)/(ab+1) is indeed a perfect square.