"Checking equality for Kolmogorov complexity of two sequences" is computable?
I think of the following argument: if we can check whether two sequences have equal Kolmogorov complexity we can write a program that enumerates all sequences of length $\le N$ and divides them into equivalence classes. 

We know that $K(x) \le |x|$. So, we have at most $N$ equivalence classes of sequences. One of this classes should be of size at least $2^{N+1}/N$. All the sequences in this class have Kolmogorov complexity at least $(N+1 - \log N) \ge N / 2$ (as we have at most $2^{n}$ sequences of Kolmogorov complexity $n$). 

So, now we have a program that can generate a sequence of Kolmogorov complexity at least $N/2$ for any given $N$. But this program itself has some fixed length $L$. Choosing $N$ to be greater than $2(L + \log N)$ leads to contradiction. 

The same argument should work for $\overset{+}{&lt;}$.

(I omited an additive constant that depends on the description language we choose to define $K(x)$.)