Calculate $\sqrt{x}$ to the power of $2$ and get $x$.
$$25x^{2}=16x$$
Subtract $16x$ from both sides.
$$25x^{2}-16x=0$$
Factor out $x$.
$$x\left(25x-16\right)=0$$
To find equation solutions, solve $x=0$ and $25x-16=0$.
$$x=0$$ $$x=\frac{16}{25}$$
Substitute $0$ for $x$ in the equation $\sqrt{x}+\sqrt{x-\sqrt{1-x}}=1$. The expression $\sqrt{0-\sqrt{1-0}}$ is undefined because the radicand cannot be negative.
$$\sqrt{0}+\sqrt{0-\sqrt{1-0}}=1$$
Substitute $\frac{16}{25}$ for $x$ in the equation $\sqrt{x}+\sqrt{x-\sqrt{1-x}}=1$.
