The expression \frac{x}{1-x} is convex on (-\infty; 1] and concave on [1; \infty).

On the convex region you rewrite \frac{x}{1-x} = \frac{1}{1-x} - 1 and represent the epigraph as t \geq \frac{1}{1-x} - 1 \Longleftrightarrow (t+1)(1-x) \geq 1,
using a rotated quadratic cone (a single second-order cone). As expected, this latter constraint implicitly bounds t+1 \geq 0 and 1-x \geq 0.

On the concave region you rewrite \frac{x}{1-x} = \frac{1}{\frac{1}{x} + \frac{1}{(-1)}} and represent the hypograph as 2t \leq \frac{2}{\frac{1}{x} + \frac{1}{(-1)}}
using the generalized harmonic mean (a single second-order cone). As expected, this latter constraint implicitly bounds x + (-1) \geq 0.

Dear hfriberg, I try to do solution but not yet solve CVX
Code given below:
cvx_begin
variable x semidefinite;
variable t semidefinite;
maximize x
subject to
(t + 1)*(1 - x) >= 1;
(t + 1) >= 0;
(1 - x) >= 0;
cvx_end