Преобразуем выражения, используя формулу квадрата суммы: $$(a + b)^2 = a^2 + 2ab + b^2$$.
- a) $$(\frac{1}{2} + a)^2 = (\frac{1}{2})^2 + 2 \cdot \frac{1}{2} \cdot a + a^2 = \frac{1}{4} + a + a^2$$
- б) $$(x + \frac{1}{3})^2 = x^2 + 2 \cdot x \cdot \frac{1}{3} + (\frac{1}{3})^2 = x^2 + \frac{2}{3}x + \frac{1}{9}$$
- в) $$(m + 0.2)^2 = m^2 + 2 \cdot m \cdot 0.2 + (0.2)^2 = m^2 + 0.4m + 0.04$$
- г) $$(1.1 + p)^2 = (1.1)^2 + 2 \cdot 1.1 \cdot p + p^2 = 1.21 + 2.2p + p^2$$
- д) $$(\frac{1}{2}a + \frac{1}{3}b)^2 = (\frac{1}{2}a)^2 + 2 \cdot \frac{1}{2}a \cdot \frac{1}{3}b + (\frac{1}{3}b)^2 = \frac{1}{4}a^2 + \frac{1}{3}ab + \frac{1}{9}b^2$$
- e) $$(\frac{3}{4}x + \frac{1}{5}y)^2 = (\frac{3}{4}x)^2 + 2 \cdot \frac{3}{4}x \cdot \frac{1}{5}y + (\frac{1}{5}y)^2 = \frac{9}{16}x^2 + \frac{3}{10}xy + \frac{1}{25}y^2$$
- ж) $$(0.2m + 2.1n)^2 = (0.2m)^2 + 2 \cdot 0.2m \cdot 2.1n + (2.1n)^2 = 0.04m^2 + 0.84mn + 4.41n^2$$
- з) $$(0.4p + 0.3q)^2 = (0.4p)^2 + 2 \cdot 0.4p \cdot 0.3q + (0.3q)^2 = 0.16p^2 + 0.24pq + 0.09q^2$$
- и) $$(\frac{3}{5}ab + \frac{1}{2}c^2)^2 = (\frac{3}{5}ab)^2 + 2 \cdot \frac{3}{5}ab \cdot \frac{1}{2}c^2 + (\frac{1}{2}c^2)^2 = \frac{9}{25}a^2b^2 + \frac{3}{5}abc^2 + \frac{1}{4}c^4$$
Ответ:
- a) $$\frac{1}{4} + a + a^2$$
- б) $$x^2 + \frac{2}{3}x + \frac{1}{9}$$
- в) $$m^2 + 0.4m + 0.04$$
- г) $$1.21 + 2.2p + p^2$$
- д) $$\frac{1}{4}a^2 + \frac{1}{3}ab + \frac{1}{9}b^2$$
- e) $$\frac{9}{16}x^2 + \frac{3}{10}xy + \frac{1}{25}y^2$$
- ж) $$0.04m^2 + 0.84mn + 4.41n^2$$
- з) $$0.16p^2 + 0.24pq + 0.09q^2$$
- и) $$\frac{9}{25}a^2b^2 + \frac{3}{5}abc^2 + \frac{1}{4}c^4$$