Question

помогите
4.16 4.17 4.18
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Answer 1

$\displaystyle\bf 1)\quad \frac{2^{50}-2\cdot4^{22}}{8^{15}\cdot31} =\frac{2^{50}-2\cdot(2^{2})^{22}}{(2^3)^{15}\cdot31} =\frac{2^{50}-2\cdot2^{44}}{2^{45}\cdot31} =\frac{2^{45}(2^5-1)}{2^{45}\cdot31} =\\\\\frac{2^{45}\cdot31}{2^{45}\cdot31} =\boxed{1} \\\\\\\ 2) \quad \frac{25^3\cdot14^2}{49\cdot10^6}=\frac{(5^2)^3\cdot(7\cdot2)^2}{7^2\cdot(5\cdot2)^6} =\frac{5^6\cdot7^2\cdot2^2}{5^6\cdot7^2\cdot2^6} =\boxed{\frac{1}{16} }$                                                                                                                                                                                                $\displaystyle\bf 3)\quad \frac{70^{40}+7^{38}-2\cdot7^{39}}{6^2\cdot49^{19}} =\frac{7^{38}(7^2+1-2\cdot7)}{7^{19\cdot2}\cdot 36} =\frac{7^{38}(49+1-14)}{7^{38}\cdot36} =\\\\\\\frac{7^{38}\cdot36}{7^{38}\cdot36} = \boxed{1}$