=8
=87a + 5b = 8a
=87a + 5b = 8a5b = 8a - 7a
=87a + 5b = 8a5b = 8a - 7aa = 5b
=87a + 5b = 8a5b = 8a - 7aa = 5b\frac{5a+7b}{b}= \frac{5*5b+7b}{b} = \frac{25b+7b}{b}= \frac{32b}{b} =32
=87a + 5b = 8a5b = 8a - 7aa = 5b\frac{5a+7b}{b}= \frac{5*5b+7b}{b} = \frac{25b+7b}{b}= \frac{32b}{b} =32 b
=87a + 5b = 8a5b = 8a - 7aa = 5b\frac{5a+7b}{b}= \frac{5*5b+7b}{b} = \frac{25b+7b}{b}= \frac{32b}{b} =32 b5a+7b
=87a + 5b = 8a5b = 8a - 7aa = 5b\frac{5a+7b}{b}= \frac{5*5b+7b}{b} = \frac{25b+7b}{b}= \frac{32b}{b} =32 b5a+7b
=87a + 5b = 8a5b = 8a - 7aa = 5b\frac{5a+7b}{b}= \frac{5*5b+7b}{b} = \frac{25b+7b}{b}= \frac{32b}{b} =32 b5a+7b =
=87a + 5b = 8a5b = 8a - 7aa = 5b\frac{5a+7b}{b}= \frac{5*5b+7b}{b} = \frac{25b+7b}{b}= \frac{32b}{b} =32 b5a+7b = b
=87a + 5b = 8a5b = 8a - 7aa = 5b\frac{5a+7b}{b}= \frac{5*5b+7b}{b} = \frac{25b+7b}{b}= \frac{32b}{b} =32 b5a+7b = b5∗5b+7b
=87a + 5b = 8a5b = 8a - 7aa = 5b\frac{5a+7b}{b}= \frac{5*5b+7b}{b} = \frac{25b+7b}{b}= \frac{32b}{b} =32 b5a+7b = b5∗5b+7b
=87a + 5b = 8a5b = 8a - 7aa = 5b\frac{5a+7b}{b}= \frac{5*5b+7b}{b} = \frac{25b+7b}{b}= \frac{32b}{b} =32 b5a+7b = b5∗5b+7b =
=87a + 5b = 8a5b = 8a - 7aa = 5b\frac{5a+7b}{b}= \frac{5*5b+7b}{b} = \frac{25b+7b}{b}= \frac{32b}{b} =32 b5a+7b = b5∗5b+7b = b
=87a + 5b = 8a5b = 8a - 7aa = 5b\frac{5a+7b}{b}= \frac{5*5b+7b}{b} = \frac{25b+7b}{b}= \frac{32b}{b} =32 b5a+7b = b5∗5b+7b = b25b+7b
=87a + 5b = 8a5b = 8a - 7aa = 5b\frac{5a+7b}{b}= \frac{5*5b+7b}{b} = \frac{25b+7b}{b}= \frac{32b}{b} =32 b5a+7b = b5∗5b+7b = b25b+7b
=87a + 5b = 8a5b = 8a - 7aa = 5b\frac{5a+7b}{b}= \frac{5*5b+7b}{b} = \frac{25b+7b}{b}= \frac{32b}{b} =32 b5a+7b = b5∗5b+7b = b25b+7b =
=87a + 5b = 8a5b = 8a - 7aa = 5b\frac{5a+7b}{b}= \frac{5*5b+7b}{b} = \frac{25b+7b}{b}= \frac{32b}{b} =32 b5a+7b = b5∗5b+7b = b25b+7b = b
=87a + 5b = 8a5b = 8a - 7aa = 5b\frac{5a+7b}{b}= \frac{5*5b+7b}{b} = \frac{25b+7b}{b}= \frac{32b}{b} =32 b5a+7b = b5∗5b+7b = b25b+7b = b32b
=87a + 5b = 8a5b = 8a - 7aa = 5b\frac{5a+7b}{b}= \frac{5*5b+7b}{b} = \frac{25b+7b}{b}= \frac{32b}{b} =32 b5a+7b = b5∗5b+7b = b25b+7b = b32b
=87a + 5b = 8a5b = 8a - 7aa = 5b\frac{5a+7b}{b}= \frac{5*5b+7b}{b} = \frac{25b+7b}{b}= \frac{32b}{b} =32 b5a+7b = b5∗5b+7b = b25b+7b = b32b =32
ответ: 1 и 2 внутренние односторонние; 3 и 4 внутренние односторонние
Пошаговое объяснение: 1 и 2 находятся на одной стороне внутри двух параллельных прямых. (Также как 3 и 4)