thanks Mark.
In fact, I need to get a lowercase x instead of a semi-positive matrix X.This is why I will write like this，

variables x(n)
X = x*x’;

As you said, this will result in a non-convex condition with a rank of 1.
Actually, the original problem of the problem is a non-convex QCQP.Convert to a SDP problem with rank 1 by defining X=x*x’.Then semi-positive relaxation, remove the rank of 1,becoming a convex SDP.

So, can I really only get matrix X? Thank you for your reply！

You can do [X x;x' 1] == semidefinite(n+1)
which by Schur Complement is equivalent to X \succeq x*x'

But because you are not otherwise using x in your formulation, I don’t believe this formulation accomplishes anything different than the formulation I provided in my previous post, which is why I didn’t mention it in that post.

The final matrix X, whose rank is not 1，but full rank.
What I know is that only when the rank of matrix X is 1, there is X=x*x’, and x is the optimal solution of the original problem.

so,as you say, i’m not otherwise using x in my formulation.Is this the reason why my problem solving failed?
But when I assume a linear item, add and do this.

But unless there are some specific properties in play here, this will not guarantee that the optimal X is rank one.

I don’t believe this formulation accomplishes anything different than the formulation I provided in my previous post

I don;t know whether there is some better convex relaxation or formulation. .But another option is to use a non-convex solver which can directly handle x*x'; this can not be done via CVX.