### 5.080 Find the largest term in the expansion of $(\sqrt{5}+\sqrt{2})^{20}$.
Solution. Let $T_{k}$ be the $k$-th term in the expansion of $(\sqrt{5}+\sqrt{2})^{20}$.

Then $\frac{T_{k+1}}{T_{k}}=\frac{C_{20}^{k} \cdot(\sqrt{5})^{20-k} \cdot(\sqrt{2})^{k}}{C_{20}^{k-1} \cdot(\sqrt{5})^{21-k} \cdot(\sqrt{2})^{k-1}}=\frac{\sqrt{2}}{\sqrt{5}} \cdot \frac{21-k}{k}$.

The ratio $\frac{T_{k+1}}{T_{k}}>1 \Leftrightarrow \frac{\sqrt{2}}{\sqrt{5}} \cdot \frac{21-k}{k}>1 \Leftrightarrow k<\frac{21\sqrt{2}}{\sqrt{2}+\sqrt{5}} \approx 8.49$.

In other words, $\left\{\begin{array}{l}T_{1}<T_{2}<\ldots<T_{9} \\ T_{9}>T_{10}>T_{11}>\ldots T_{21} .\end{array}\right.$

Answer: The 9th term.