296 3) log24 + log2 √10 log2 20 + 3 log22

$$\frac{log_2 4 + log_2 \sqrt{10}}{log_2 20 + 3log_2 2} = \frac{log_2 (4\sqrt{10})}{log_2 20 + log_2 2^3} = \frac{log_2 (4\sqrt{10})}{log_2 20 + log_2 8} = \frac{log_2 (4\sqrt{10})}{log_2 (20 \cdot 8)} = \frac{log_2 (4\sqrt{10})}{log_2 160} = \frac{log_2 (2^2 \cdot 10^{\frac{1}{2}})}{log_2 (16 \cdot 10)} = \frac{log_2 2^2 + log_2 10^{\frac{1}{2}}}{log_2 2^4 + log_2 10} = \frac{2log_2 2 + \frac{1}{2}log_2 10}{4log_2 2 + log_2 10} = \frac{2 + \frac{1}{2}log_2 10}{4 + log_2 10} = \frac{\frac{1}{2}(4 + log_2 10)}{4 + log_2 10} = \frac{1}{2}$$.

Ответ: 1/2