\(\dfrac{x+3}{4}=\dfrac{16}{x+3}\)
\(\Rightarrow\left(x+3\right)\left(x+3\right)=16.4\)
\(\Rightarrow\left(x+3\right)^2=64=8^2\)
\(\Rightarrow x+3=\pm8\)
\(\Rightarrow x+3\in\left\{-8;8\right\}\Rightarrow x\in\left\{-11;5\right\}\)
Vậy \(x\in\left\{-11;5\right\}\)