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+### Statement
+
+$5.3.9.$ [Insert the problem statement]
+
+### Solution
+
+
+We'll consider molecules $A$ as impurities, meaning that the number of collisions of molecules $A$ with each other is much less than the number of collisions with molecules of gases $B_{1}$ and $B_{2}$. We'll also assume that the temperature is the same in all cases. This means we can assume that:
+
+$$\frac{D_{1}}{D_{2}}=\frac{(R_{A} + R_{B_{2}})^{2}}{(R_{A} + R_{B_{1}})^{2}}$$
+
+
+Number of collisions in the mixture:
+
+$$\nu_{12}=\pi\cdot (R_{A} + R_{B_{1}})^{2}\cdot n_{1}\cdot <\upsilon_{sq_{A}}> + \pi\cdot (R_{A} + R_{B_{2}})^{2}\cdot n_{2}\cdot <\upsilon_{sq_{A}}>$$
+
+
+Let's write down the ratio of $D_{1}$ to $D_{12}$, after some transformations we get:
+
+$$\frac{D_{1}}{D_{12}}=\frac{(R_{A} + R_{B_{1}})^{2}\cdot n_{1} + (R_{A} + R_{B_{2}})^{2}\cdot n_{2}}{(R_{A} + R_{B_{1}})^{2}\cdot n}$$
+
+
+Let's use the relation for $D_{1}$ and $D_{2}$:
+
+$$\frac{D_{1}}{D_{12}}=\frac{n_{1} + \frac{D_{1}}{D_{2}}\cdot n_{2}}{n}$$
+
+
+Expressing $D_{12}$ we get:
+
+$$D_{12}=\frac{n\cdot D_{1}\cdot D_{2}}{n_{1}\cdot D_{2} + n_{2}\cdot D_{1}}$$
+
+#### Answer
+
+[Insert a concise answer or boxed result]