[tex]\begin{gathered} (f+g)(x)=f(x)+g(x) \\ =(x^2+3)+\sqrt[]{5-x} \\ =x^2+\sqrt[]{5-x}+3 \\ \text{ So we have} \\ (f+g)(x)=x^2+\sqrt[]{5-x}+3 \end{gathered}[/tex][tex]\begin{gathered} (f-g)(x)=f(x)-g(x) \\ =(x^2+3)-\sqrt[]{5-x} \\ =x^2-\sqrt[]{5-x}+3 \\ \text{thus } \\ (f-g)(x)=x^2-\sqrt[]{5-x}+3 \end{gathered}[/tex][tex]\begin{gathered} (fg)(x)=f(x)g(x) \\ =(x^2+3)\sqrt[]{5-x} \\ =x^2\sqrt[]{5-x}+3\sqrt[]{5-x} \\ \text{ then we get} \\ (fg)(x)=x^2\sqrt[]{5-x}+3\sqrt[]{5-x} \end{gathered}[/tex][tex]\begin{gathered} (\frac{f}{g})(x)=\frac{f(x)}{g(x)} \\ =\frac{x^2+3}{\sqrt[]{5-x}} \\ =\frac{x^2+3}{\sqrt[]{5-x}}\frac{\sqrt[]{5-x}}{\sqrt[]{5-x}} \\ =\frac{(x^2+3)\sqrt[]{5-x}}{(\sqrt[]{5-x})^2} \\ =\frac{x^2\sqrt[]{5-x}+3\sqrt[]{5-x}}{5-x} \\ \text{thus } \\ (\frac{f}{g})(x)=\frac{x^2\sqrt[]{5-x}+3\sqrt[]{5-x}}{5-x} \end{gathered}[/tex]
