[x].

Let $x=n+k$ , where $n$ is an integer and $0\leq k<1$ . Therefore $\lfloor n+k\rfloor=n$ , so $\frac2{n+k}=\frac1n+\frac1{\lfloor2n+2k\rfloor}$ .

Because of the $\lfloor2x\rfloor$ , there are two possible options: $\begin{aligned}0\leq\,&k<\frac12:&\frac12\leq\,&k<1:\\ \lfloor2n+2k\rfloor&=2n&\lfloor2n+2k\rfloor&=2n+1\\ \therefore\frac2{n+k}&=\frac1n+\frac1{2n}&\therefore\frac2{n+k}&=\frac1n+\frac1{2n+1}\\ \frac2{n+k}&=\frac3{2n}&2n(2n+1)&=(n+k)(2n+1)+n(n+k)\\ 4n&=3n+3k&4n^2+2n&=2n^2+n+2kn+k+n^2+kn\\ n&=3k&n^2+n&=3kn+k\\ k&=\frac n3&k&=\frac{n^2+n}{3n+1}\end{aligned}$ The first option is only possible if $n=1$ , as $k$ must be less than $\frac12$ . This means $x=1.333...$

For the second option, we can rearrange the expression to be $k=\frac n3+\frac29-\frac2{9(3n+1)}$ , so that as $n$ approaches infinity $k$ approaches $\frac n3+\frac29$ from the negative side for $n>0$ . Since $\frac n3+\frac29$ is increasing, $\frac n3+\frac29-\frac2{9(3n+1)}$ must also be increasing for $n>0$ .

This means the largest possible value of $n$ will be the largest integer before $k$ exceeds 1, as for this option $\frac12\leq k<1$ . Through experimentation, we find that this value is $n=2$ , as when $n=3, k>1$ . From the equation above, when $n=2, k=\frac67$ , which means the largest positive value of x is $2+\frac67\approx\boxed{2.8571}$