And thanks in advance for your answers. Sorry if the text is badly formatted, I'm new here. Anyway, here is the question:
Find all real numbers x, y, z, u and v in $\sqrt{x}+\sqrt{y}+2\sqrt{z-2}+\sqrt{u}+\sqrt{v}=x+y+z+u+v$
Could I use the method of completing the square (I'm familiar with that) or is there some better way?
On
Yes, you can solve your equation by completing the squares. $$\sqrt{x}+\sqrt{y}+2\sqrt{z-2}+\sqrt{u}+\sqrt{v}=x+y+z+u+v\tag{*1}$$ For each variable, subtract its appearance in LHS from corresponding term in RHS, you have: $$ \begin{array}{rl} x - \sqrt{x} &= (\sqrt{x}-\frac12)^2 - \frac14\\ y - \sqrt{y} &= (\sqrt{y}-\frac12)^2 - \frac14\\ z - 2\sqrt{z-2} &= (\sqrt{z-2}-1)^2 + 1\\ u - \sqrt{u} &= (\sqrt{u}-\frac12)^2 - \frac14\\ v - \sqrt{v} &= (\sqrt{v}-\frac12)^2 - \frac14\\ \end{array} $$ Now sum over both sides and compare result with $(*1)$, you find: $$\verb/RHS/(*1) - \verb/LHS/(*1) = (\sqrt{x}-\frac12)^2 + (\sqrt{y}-\frac12)^2 + (\sqrt{z-2}-1)^2 + (\sqrt{u}-\frac12)^2 + (\sqrt{v}-\frac12)^2$$ This leads to $$\begin{cases} \sqrt{x} = \sqrt{y} = \sqrt{u} = \sqrt{v} = \frac12,\\ \sqrt{z-2} = 1 \end{cases} \quad\implies\quad (x,y,z,u,v) = \left(\frac14,\frac14,3,\frac14,\frac14\right)$$
HINT.-You have always $2\sqrt{z-2}-z\le -1$ and $x-\sqrt x$ takes its minumun at $x=\frac 14$ this minimun being equal to $-\frac 14$. Since $$2\sqrt{z-2}-z=(x-\sqrt x)+(y-\sqrt y)+(u-\sqrt u)+(v-\sqrt v)$$ it follows that the only solution is $$(x,y,z,u,v)=(\frac14,\frac14,3,\frac14,\frac14)$$