semi decidability: there are two languages ​​$ L_1 $ and $ L_2 $ so that $ L_1 $ and $ L_2 $ are recursive, but $ L_1L_2 $ is not recursive

True or false statements.

There are two languages $ L_1 $ Y $ L_2 $ such that $ L_1 $ Y $ L_2 $ are recurvise but $ L_1L_2 $ It is not recursive.

For me, the statements are true because I can describe a non-recursive one (I understood recursively enumerable) procedure of two recurring languages.

Test:

Suppose that $ L_1 $ Y $ L_2 $ they are recursive languages ​​Then there is a Turing M1 machine that accepts any input in $ L_1 $ and reject any entry that is not in $ L_1 $. Similarly, there is a Turing M2 machine that accepts any input in $ L_2 $ and reject any entry that is not in $ L_2 $. To determine if $ w ∈ L_1L_2 $, we guess in a non-determining way where to divide the chain into $ w_1w_2 $and run M1 in $ w_1 $ and M2 in $ w_2 $. So, if w is in the language, M1 and M2 will eventually accept.

Does it involve $ L_1L_2 = L_2L_1 $ $ L_1 = L_2?

Leave $ L_1, L_2 subseteq Sigma ^ * $ be two languages, where $ Sigma $ It is a finite alphabet.

Make $ L_1L_2 = L_2L_1 $ to imply $ L_1 = L_2 $?

What if $ L_1 $ Y $ L_2 $ Are they regular languages?

Can you give counterexamples?

demonstrating L1 * ∪ L2 * ⊆ (L1∪L2) *

x∈ L1 * ∪ L2 * ⇔ x∈ L1 * ∨ x ∈L2 * ⇔ x ∈ (L1) * ∨ x∈ (L2) * ⇔ x ∈L1 * ∪ L2 * ⇔ x∈ (L1∪L2) *

Is it enough to try it this way?