The formula for the binomial expansion of (1+ax)^n, where  n is negative is:
1+n(ax)+(n⋅(n−1)/2!)*(ax)^2...((n(n-1)...(n−r+1))/r!)(ax)^r.

As x^2 and higher powers can be neglected, (1+ax)^n = 1+n.ax

So, the given expression equates to ((7/4)^(1/2))*(1+4x/3) which is equal to (root(7)/2)*(1+4x/3).

Hope it helps.