G7.3 Find the largest value of $c$, if $c=2-x+2 \sqrt{x-1}$ and $x>1$.
$$
\begin{array}{l}
(c+x-2)^{2}=4(x-1) \\
c^{2}+x^{2}+4+2 c x-4 x-4 c=4 x-4 \\
x^{2}+2(c-4) x+\left(c^{2}-4 c+8\right)=0
\end{array}
$$

For real values of $x, \Delta \geq 0$
$$
\begin{array}{l}
4(c-4)^{2}-4\left(c^{2}-4 c+8\right) \geq 0 \\
c^{2}-8 c+16-c^{2}+4 c-8 \geq 0 \\
8 \geq 4 c \\
\Rightarrow c \leq 2
\end{array}
$$
$\Rightarrow$ The largest value $c=2$
Method 2
Let $y=\sqrt{x-1}$, then $y^{2}=x-1$
$$
\begin{array}{l}
\Rightarrow x=y^{2}+1 \\
c=2-\left(y^{2}+1\right)+2 y=2-(1-y)^{2} \leq 2
\end{array}
$$

The largest value of $c=2$.