1. Solve the equation $\sqrt[3]{2-x}+\sqrt{1-x}=1$.
Solution. Introducing the substitutions $y=\sqrt[3]{2-x}$ and $z=\sqrt{x-1}$, we obtain the system of equations $y^{3}+x=2, z^{2}+1=x, y+z=1$. From the first and second equations, we get
$y^{3}+z^{2}=1$. Substituting $z=1-y$, we get $y\left(y^{2}+y-2\right)=0$. The solutions to $y^{3}+y^{2}-2 y=0$ are $y_{1}=0$ and $y_{2 / 3}=\frac{-1 \pm \sqrt{1+8}}{2}$, i.e., $y_{1}=0, y_{2}=1, y_{3}=-2$. From $x=2-y^{3}$, the desired solutions are $x_{1}=0, x_{2}=1, x_{3}=10$.