How to find x from the equation $\sqrt{x + \sqrt{x}} - \sqrt{x - \sqrt{x}} = m \sqrt{\frac{x}{x+\sqrt{x}}}$ For m are real number,
Find x from the equation
$$\sqrt{x + \sqrt{x}} - \sqrt{x - \sqrt{x}} = m \sqrt{\frac{x}{x+\sqrt{x}}}$$
I tried to multiply $\sqrt{x + \sqrt{x}}$ to the both sides and I get
$$x + \sqrt{x} - \sqrt{x^{2} - x} = m \sqrt{x}$$
What should I do now to get x? Can anyone show me a hint please?

From the simplified form of @gt6989b, we get $$\sqrt{x}- \sqrt{x-1}=m-1.$$ 
Squaring for rationalization one must demand that $$m \ge 1...(1),~~ x \ge 1....(2),~~~\sqrt{x} \ge m-1....(3)$$.The Eq. (3) gives a very interestiong condition which has been missed out in the answers by @Dinno Koluh  and @trancelocation. When we demand (1) and (3), we get
$$\sqrt{x} \ge \frac{m^2-2m+2}{2(m-1)} \ge (m-1) \implies m^2-2m \le 0 \implies m(m-2) \le 0 \implies 1<m \le 2.$$ Finally, we clam that the solution of the priposed equation is given as
$$x=\left(\frac{m^2-2m+2}{2(m-1)}\right)^2  ~~if~~ 1<m \le 2.$$
