5.081 For what values of $x$ is the fourth term the largest in the expansion of $(5+3 x)^{10}$?
Solution. Let $T_{k}$ be the $k$-th term in the expansion of $(5+3 x)^{10}$.

Then $T_{4}=C_{10}^{3} \cdot 5^{7} \cdot(3 x)^{3}$ and the maximum of $T_{4}$ is reached when $x>0$.

We calculate $\frac{T_{k}}{T_{k-1}}=\frac{C_{10}^{k-1} \cdot 5^{11-k} \cdot(3 x)^{k-1}}{C_{10}^{k-2} \cdot 5^{12-k} \cdot(3 x)^{k-2}}=\frac{2 x}{5} \cdot \frac{12-k}{k-1}, k \geq 2$.

From this, $\frac{T_{k}}{T_{k-1}} \geq 1 \Leftrightarrow k \leq \frac{36 x+5}{3 x+5}, k \geq 2$.

Thus, $\left\{\begin{array}{l}T_{2} \leq T_{3} \leq \ldots T_{k} \\ T_{k}>T_{k+1}>\ldots T_{11}\end{array}\right.$ when $k=\left[\frac{36 x+5}{3 x+5}\right]$, where [a] denotes the integer part of the number $a$.

By the condition, $T_{4}$ is the largest term in the expansion $\Leftrightarrow k=4$ and $T_{3}120 \cdot 5^{7} \cdot 3^{3} \cdot\left(\frac{5}{8}\right)^{3}>T_{1}=5^{10}$.

Answer: $x \in\left(\frac{5}{8}, \frac{20}{21}\right)$