$\left(x+\dfrac{1}{x}\right)^2=x^2+\dfrac{1}{x^2}-2=9\implies x+\dfrac{1}{x}=3 \\ \text{Ta có:} \ x^3+\dfrac{1}{x^3}=\left(x^2+\dfrac{1}{x^2}\right)\left(x+\dfrac{1}{x}\right)-\left(x+\dfrac{1}{x}\right)=7.3-3=18 \\ x^4+\dfrac{1}{x^4}=\left(x^2+\dfrac{1}{x^2}\right)^2-2=49-2=47\\ \implies x^5+\dfrac{1}{x^5}=\left(x^4+\dfrac{1}{x^4}\right)\left(x+\dfrac{1}{x}\right)-\left(x^3+\dfrac{1}{x^3}\right)=47.3-18=\boxed{123}$