<s>[INST] Question on binomial theorem: Find the term independent of $x$ in the expansion of $(2+x)\left(2x+\frac{1}{x}\right)^5$ 
Find the term independent of $x$ in the expansion of $$(2+x)\left(2x+\frac{1}{x}\right)^5$$

I am only able to do this with one binomial, but when there are two I do not totally understand the process to do this.
 [/INST] You can make life a bit easier in first writing
$$(2+x)\left(2x+\frac 1x\right)^5 = \frac 1{x^5}\color{blue}{(2+x)(2x^2+1)^5}$$
Now, you see you need only calculate the coefficient of $x^5$ in $(2+x)(2x^2+1)^5$. This is easy since $(2x^2+1)^5$ contains only even powers of $x$ but you are looking for $x^5$. Hence using binomial expansion you get
\begin{eqnarray*}[x^5](2+x)(2x^2+1)^5
& = & [x^5](2+x)\binom 52(2x^2)^2 \\
& = & \binom 52 \cdot 4 = 40
\end{eqnarray*}
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