what is the constant term in the expansion of \(\left(\sqrt{x}+\dfrac5x\right)^{9}\)?
+36915 WolframAlpha.com says 10500 but I do not know how to get there ! 
Let's see what Melody has to say.....
+118608 the general term is
\(T_n=\binom {9}{n} \left(\frac{5}{x}\right)^n(x^{0.5})^{9-n}\\ T_n=\binom {9}{n} (5^n)(x^{-n})(x^{4.5-0.5n})\\ T_n=\binom {9}{n} (5^n)(x^{4.5-1.5n})\\ \text{for the constant term} \quad \\ 4.5-1.5n=0\\ 9-3n=0\\ 3n=9\\ n=3\\ T_3=\binom {9}{3} (5^3)\\ T_3=84 *125\\ T_3=10500 \)
You need to check this.
Coding:
T_n=\binom {9}{n} \left(\frac{5}{x}\right)^n(x^{0.5})^{9-n}\\
T_n=\binom {9}{n} (5^n)(x^{-n})(x^{4.5-0.5n})\\
T_n=\binom {9}{n} (5^n)(x^{4.5-1.5n})\\
\text{for the constant term} \quad \\
4.5-1.5n=0\\
9-3n=0\\
3n=9\\
n=3\\
T_3=\binom {9}{3} (5^3)\\
T_3=84 *125\\
T_3=10500
