Ответ:
Пояснение к Заданию 4:
Все выражения под корнем являются полными квадратами:
- \[ \sqrt{a^2+8ab+16b^2} = \sqrt{(a+4b)^2} = |a+4b| \]
- \[ \sqrt{a^2+12ab+36b^2} = \sqrt{(a+6b)^2} = |a+6b| \]
- \[ \sqrt{a^2+10ab+25b^2} = \sqrt{(a+5b)^2} = |a+5b| \]
- \[ \sqrt{9a^2+6ab+b^2} = \sqrt{(3a+b)^2} = |3a+b| \]
- \[ \sqrt{16a^2+8ab+b^2} = \sqrt{(4a+b)^2} = |4a+b| \]
- \[ \sqrt{25a^2+10ab+b^2} = \sqrt{(5a+b)^2} = |5a+b| \]
- \[ \sqrt{36a^2+12ab+b^2} = \sqrt{(6a+b)^2} = |6a+b| \]
- \[ \sqrt{a^2-6ab+9b^2} = \sqrt{(a-3b)^2} = |a-3b| \]
- \[ \sqrt{a^2-12ab+36b^2} = \sqrt{(a-6b)^2} = |a-6b| \]
- \[ \sqrt{a^2-8ab+16b^2} = \sqrt{(a-4b)^2} = |a-4b| \]
- \[ \sqrt{a^2-10ab+25b^2} = \sqrt{(a-5b)^2} = |a-5b| \]
- \[ \sqrt{a^2+4ab+4b^2} = \sqrt{(a+2b)^2} = |a+2b| \]
Теперь подставим значения a и b в каждое выражение.
Пошаговое решение:
Задание 4
- \[ |3 + 4 \cdot \frac{1}{7}| = |3 + \frac{4}{7}| = |\frac{21+4}{7}| = \frac{25}{7} \]
- \[ |\frac{7}{5} + 6 \cdot \frac{3}{5}| = |\frac{7}{5} + \frac{18}{5}| = |\frac{25}{5}| = 5 \]
- \[ |\frac{16}{13} + 5 \cdot \frac{4}{13}| = |\frac{16}{13} + \frac{20}{13}| = |\frac{36}{13}| = \frac{36}{13} \]
- \[ |3\frac{2}{3} + 4 \cdot \frac{1}{3}| = |\frac{11}{3} + \frac{4}{3}| = |\frac{15}{3}| = 5 \]
- \[ |3 \cdot \frac{5}{13} + \frac{11}{13}| = |\frac{15}{13} + \frac{11}{13}| = |\frac{26}{13}| = 2 \]
- \[ |4 \cdot \frac{3}{11} + \frac{10}{11}| = |\frac{12}{11} + \frac{10}{11}| = |\frac{22}{11}| = 2 \]
- \[ |5 \cdot \frac{4}{9} + \frac{7}{9}| = |\frac{20}{9} + \frac{7}{9}| = |\frac{27}{9}| = 3 \]
- \[ |6 \cdot \frac{4}{5} + \frac{8}{5}| = |\frac{24}{5} + \frac{8}{5}| = |\frac{32}{5}| = \frac{32}{5} \]
- \[ |3 - 3 \cdot 6| = |3 - 18| = |-15| = 15 \]
- \[ |8 - 6 \cdot 3| = |8 - 18| = |-10| = 10 \]
- \[ |4 - 4 \cdot 3| = |4 - 12| = |-8| = 8 \]
- \[ |7 - 5 \cdot 2| = |7 - 10| = |-3| = 3 \]
- \[ |8 + 5 \cdot (-2)| = |8 - 10| = |-2| = 2 \]
- \[ |5 + 3 \cdot (-4)| = |5 - 12| = |-7| = 7 \]
- \[ |7 + 6 \cdot (-3)| = |7 - 18| = |-11| = 11 \]
- \[ |2 + 2 \cdot (-4)| = |2 - 8| = |-6| = 6 \]
