How to derive easily : $|\sqrt{x+3} -2|=\frac {|x-1|}{\sqrt{x+3}+2}$? ( Context : a limit problem). Source : Lebossé & Hemery, Algèbre et Analyse ( Terminales CDT ), $1966$.
$|\sqrt{x+3} -2|=\frac {|x-1|}{\sqrt{x+3}+2}$
This equality is asserted in the solution of a limit problem , namely : show that the limit of $f(x)=\sqrt{x+3}$ when $x$ goes to $1$  is $2$.
I tried to understand this assertion using the fact that $a/b = c/d$ implies that $ a\times c = b\times d$.
So I said to myself that $|\sqrt{x+3} -2|\times (\sqrt {x+3} +2)$ should yield  $|x+1|$, and that , in that case , the assertion would be justified.
Since $\sqrt{x+3} -2 \geq 0 \iff x\geq 1$ one has
(1) if $x\geq 1 $ , $|\sqrt{x+3} -2|= \sqrt{x+3} -2$ , hence $|\sqrt{x+3} -2|\times (\sqrt {x+3} +2)= (\sqrt{x+3} -2)\times (\sqrt {x+3} +2) = (\sqrt{x+3})^2 - ( 2^2) = (x+3) - 4 = x-1$
(2) if $x\lt 1 $ , $|\sqrt{x+3} -2|= -(\sqrt{x+3} -2) = 2- \sqrt{x+3}$ , hence $|\sqrt{x+3} -2|\times (\sqrt {x+3} +2)= 2- \sqrt{x+3} \times (2+ \sqrt {x+3}) = (2^2 - (\sqrt{x+3})^2 = 4- (x+3) = 4-x-3 = 1-x $.
My question :
(1) how do we go formally from " the product is $(x-1)$ if $x\geq 1$ and $(1-x)$ if $x\lt 1$ to " the product is equal to $|x-1|$" ? I see that $x-1$ is the ( additive ) inverse of $1-x$ but I cannot use it to draw the desired conclusion .
(2) is there a quick way to arrive at $|\sqrt{x+3} -2|=\frac {|x-1|}{\sqrt{x+3}+2}$, without all these manipulations?

Regardless of the value of $x$, $\sqrt{x+3}+2>0$, so $\sqrt{x+3}+2=|\sqrt{x+3}+2|$ and $(\sqrt{x+3}+2)^{-1}$ exists.
Multiplying $(\sqrt{x+3}+2)^{-1}|\sqrt{x+3}+2|=1$, $|\sqrt{x+3}-2|$ changes into...
$$
|\sqrt{x+3}-2|=\frac{|\sqrt{x+3}-2|\cdot|\sqrt{x+3}+2|}{\sqrt{x+3}+2}=\frac{|\sqrt{x+3}^2-2^2|}{\sqrt{x+3}+2}=\frac{|x-1|}{\sqrt{x+3}+2}
$$
