[tex]f(x)=\dfrac{1}{x^2-3}\\\\a)\ f(3)\to\text{put x=3 to the equation of the function f}\\\\f(3)=\dfrac{1}{3^2-3}=\dfrac{1}{9-3}=\dfrac{1}{6}\\\\Answer:\ \boxed{f(3)=\dfrac{1}{6}}\\\\b)\ f(2+h)\to\text{put x=2+h to the equation of a function f}\\\\f(2+h)=\dfrac{1}{(2+h)^2-3}\ \ \ |\text{use}\ (a+b)^2=a^2+2ab+b^2\\\\=\dfrac{1}{2^2+2(2)(h)+h^2-3}=\dfrac{1}{4+4h+h^2-3}=\dfrac{1}{h^2+4h+1}\\\\Answer:\ \boxed{f(2+h)=\dfrac{1}{h^2+4h+1}}[/tex]
