Найди значение выражения √a2 (-а)2 при а=4

Конечно, давай решим эти задания по порядку! **Задание 3** 1. $\sqrt{a^2} \cdot (-a)^2$ при $a = 4$ * $\sqrt{4^2} \cdot (-4)^2 = \sqrt{16} \cdot 16 = 4 \cdot 16 = 64$ 2. $\sqrt{a^6} \cdot (-a)^4$ при $a = 2$ * $\sqrt{2^6} \cdot (-2)^4 = \sqrt{64} \cdot 16 = 8 \cdot 16 = 128$ 3. $\sqrt{a^6} \cdot (-a)^2$ при $a = 3$ * $\sqrt{3^6} \cdot (-3)^2 = \sqrt{729} \cdot 9 = 27 \cdot 9 = 243$ 4. $\sqrt{a^2} \cdot (-a)^4$ при $a = 4$ * $\sqrt{4^2} \cdot (-4)^4 = \sqrt{16} \cdot 256 = 4 \cdot 256 = 1024$ **Задание 4** 1. $\sqrt{a^2 + 8ab + 16b^2}$ при $a = 3\frac{3}{7}, b = \frac{1}{7}$; * $\sqrt{(3\frac{3}{7})^2 + 8 \cdot 3\frac{3}{7} \cdot \frac{1}{7} + 16 \cdot (\frac{1}{7})^2} = \sqrt{(\frac{24}{7})^2 + 8 \cdot \frac{24}{7} \cdot \frac{1}{7} + 16 \cdot \frac{1}{49}} = \sqrt{\frac{576}{49} + \frac{192}{49} + \frac{16}{49}} = \sqrt{\frac{784}{49}} = \sqrt{16} = 4$ 2. $\sqrt{a^2 + 12ab + 36b^2}$ при $a = 7\frac{2}{5}, b = \frac{3}{5}$; * $\sqrt{(7\frac{2}{5})^2 + 12 \cdot 7\frac{2}{5} \cdot \frac{3}{5} + 36 \cdot (\frac{3}{5})^2} = \sqrt{(\frac{37}{5})^2 + 12 \cdot \frac{37}{5} \cdot \frac{3}{5} + 36 \cdot \frac{9}{25}} = \sqrt{\frac{1369}{25} + \frac{1332}{25} + \frac{324}{25}} = \sqrt{\frac{3025}{25}} = \sqrt{121} = 11$ 3. $\sqrt{a^2 + 10ab + 25b^2}$ при $a = 1\frac{6}{13}, b = \frac{4}{13}$; * $\sqrt{(1\frac{6}{13})^2 + 10 \cdot 1\frac{6}{13} \cdot \frac{4}{13} + 25 \cdot (\frac{4}{13})^2} = \sqrt{(\frac{19}{13})^2 + 10 \cdot \frac{19}{13} \cdot \frac{4}{13} + 25 \cdot \frac{16}{169}} = \sqrt{\frac{361}{169} + \frac{760}{169} + \frac{400}{169}} = \sqrt{\frac{1521}{169}} = \sqrt{9} = 3$ 4. $\sqrt{a^2 + 8ab + 16b^2}$ при $a = 3\frac{2}{3}, b = \frac{1}{3}$; * $\sqrt{(3\frac{2}{3})^2 + 8 \cdot 3\frac{2}{3} \cdot \frac{1}{3} + 16 \cdot (\frac{1}{3})^2} = \sqrt{(\frac{11}{3})^2 + 8 \cdot \frac{11}{3} \cdot \frac{1}{3} + 16 \cdot \frac{1}{9}} = \sqrt{\frac{121}{9} + \frac{88}{9} + \frac{16}{9}} = \sqrt{\frac{225}{9}} = \sqrt{25} = 5$ 5. $\sqrt{9a^2 + 6ab + b^2}$ при $a = \frac{5}{13}, b = 6\frac{11}{13}$; * $\sqrt{9 \cdot (\frac{5}{13})^2 + 6 \cdot \frac{5}{13} \cdot 6\frac{11}{13} + (6\frac{11}{13})^2} = \sqrt{9 \cdot \frac{25}{169} + 6 \cdot \frac{5}{13} \cdot \frac{89}{13} + (\frac{89}{13})^2} = \sqrt{\frac{225}{169} + \frac{2670}{169} + \frac{7921}{169}} = \sqrt{\frac{10816}{169}} = \sqrt{64} = 8$ 6. $\sqrt{16a^2 + 8ab + b^2}$ при $a = \frac{3}{11}, b = 5\frac{10}{11}$; * $\sqrt{16 \cdot (\frac{3}{11})^2 + 8 \cdot \frac{3}{11} \cdot 5\frac{10}{11} + (5\frac{10}{11})^2} = \sqrt{16 \cdot \frac{9}{121} + 8 \cdot \frac{3}{11} \cdot \frac{65}{11} + (\frac{65}{11})^2} = \sqrt{\frac{144}{121} + \frac{1560}{121} + \frac{4225}{121}} = \sqrt{\frac{5929}{121}} = \sqrt{49} = 7$ 7. $\sqrt{25a^2 + 10ab + b^2}$ при $a = \frac{4}{9}, b = 3\frac{7}{9}$; * $\sqrt{25 \cdot (\frac{4}{9})^2 + 10 \cdot \frac{4}{9} \cdot 3\frac{7}{9} + (3\frac{7}{9})^2} = \sqrt{25 \cdot \frac{16}{81} + 10 \cdot \frac{4}{9} \cdot \frac{34}{9} + (\frac{34}{9})^2} = \sqrt{\frac{400}{81} + \frac{1360}{81} + \frac{1156}{81}} = \sqrt{\frac{2916}{81}} = \sqrt{36} = 6$ 8. $\sqrt{36a^2 + 12ab + b^2}$ при $a = \frac{4}{5}, b = 8\frac{1}{5}$. * $\sqrt{36 \cdot (\frac{4}{5})^2 + 12 \cdot \frac{4}{5} \cdot 8\frac{1}{5} + (8\frac{1}{5})^2} = \sqrt{36 \cdot \frac{16}{25} + 12 \cdot \frac{4}{5} \cdot \frac{41}{5} + (\frac{41}{5})^2} = \sqrt{\frac{576}{25} + \frac{1968}{25} + \frac{1681}{25}} = \sqrt{\frac{4225}{25}} = \sqrt{169} = 13$ Надеюсь, теперь тебе стало понятнее!