algebra binomial help

Hi proyaop,

The general term is

$\displaystyle {\binom{9}{n}(\sqrt x)^{9-n}(\frac{5}{x})^n}\\~\\ =\displaystyle {\binom{9}{n}(x)^{\frac{9-n}{2}}(\frac{5}{x^n})}\\~\\ =\displaystyle {\binom{9}{n}*5(x)^{\frac{9-n-2n}{2}}}\\~\\ =\displaystyle {\binom{9}{n}*5(x)^{\frac{9-3n}{2}}}\\~\\$

You want the constant term so     

$\frac{9-3x}{2}=0\\ 9-3x=0\\ 9=3x\\ x=3$

So the contatn term is

$=\displaystyle {\binom{9}{n}*5(x)^{\frac{9-3n}{2}}}\\~\\ =\displaystyle {\binom{9}{3}*5(x)^{\frac{9-3*3}{2}}}\\~\\ =\binom{9}{3}*5\\~\\ =84*5\\~\\ =420$

You do need to check my working for careless mistakes of course. 

LaTex:

\displaystyle {\binom{9}{n}(\sqrt x)^{9-n}(\frac{5}{x})^n}\\~\\
=\displaystyle {\binom{9}{n}(x)^{\frac{9-n}{2}}(\frac{5}{x^n})}\\~\\
=\displaystyle {\binom{9}{n}*5(x)^{\frac{9-n-2n}{2}}}\\~\\
=\displaystyle {\binom{9}{n}*5(x)^{\frac{9-3n}{2}}}\\~\\