Answer
$a)$ $(f+g)(x)=\sqrt{x}+x+5$
$b)$ $(f-g)(x)=\sqrt{x}-x-5$
$c)$ $(f\cdot g)(x)=x\sqrt{x}+5\sqrt{x}$
$d)$ $\Big(\dfrac{f}{g}\Big)(x)=\dfrac{\sqrt{x}}{x+5}$, where $x\ne-5$
Work Step by Step
$f(x)=\sqrt{x};$ $g(x)=x+5$
For each case, replace $f(x)$ by $\sqrt{x}$ and $g(x)$ by $x+5$ and simplify if possible:
$a)$ $(f+g)(x)$
$f(x)+g(x)=\sqrt{x}+x+5=\sqrt{x}+x+5$
$b)$ $(f-g)(x)$
$f(x)-g(x)=(\sqrt{x})-(x+5)=\sqrt{x}-x-5$
$c)$ $(f\cdot g)(x)$
$f(x)\cdot g(x)=(\sqrt{x})(x+5)=x\sqrt{x}+5\sqrt{x}$
$d)$ $\Big(\dfrac{f}{g}\Big)(x)$
$\dfrac{f(x)}{g(x)}=\dfrac{\sqrt{x}}{x+5}$, where $x\ne-5$