Представим многочлены в виде произведения, используя формулу разности квадратов: $$a^2 - b^2 = (a-b)(a+b)$$.
- а) $$x^2 - 36 = x^2 - 6^2 = (x-6)(x+6)$$
- б) $$1 - a^2 = 1^2 - a^2 = (1-a)(1+a)$$
- в) $$16 - x^2 = 4^2 - x^2 = (4-x)(4+x)$$
- г) $$-y^2 + 81 = 81 - y^2 = 9^2 - y^2 = (9-y)(9+y)$$
- д) $$\frac{1}{9} - b^2 = (\frac{1}{3})^2 - b^2 = (\frac{1}{3}-b)(\frac{1}{3}+b)$$
- е) $$y^2 - \frac{25}{36} = y^2 - (\frac{5}{6})^2 = (y - \frac{5}{6})(y + \frac{5}{6})$$
- ж) $$0.81 - x^2 = (0.9)^2 - x^2 = (0.9 - x)(0.9 + x)$$
- з) $$16a^2 - 1 = (4a)^2 - 1^2 = (4a - 1)(4a + 1)$$
- и) $$100 - 9y^2 = 10^2 - (3y)^2 = (10 - 3y)(10 + 3y)$$
- к) $$36a^2 - 25b^2 = (6a)^2 - (5b)^2 = (6a - 5b)(6a + 5b)$$
- л) $$-9p^2 + 0.16q^2 = 0.16q^2 - 9p^2 = (0.4q)^2 - (3p)^2 = (0.4q - 3p)(0.4q + 3p)$$
- м) $$\frac{4}{81}k^2 - \frac{1}{25}b^2 = (\frac{2}{9}k)^2 - (\frac{1}{5}b)^2 = (\frac{2}{9}k - \frac{1}{5}b)(\frac{2}{9}k + \frac{1}{5}b)$$
- н) $$7\frac{1}{9}n^2 - 4m^2 = (\sqrt{7\frac{1}{9}}n)^2 - (2m)^2 = (\sqrt{\frac{64}{9}}n - 2m)(\sqrt{\frac{64}{9}}n + 2m) = (\frac{8}{3}n - 2m)(\frac{8}{3}n + 2m)$$
- о) $$0.04x^2 - 0.64y^2 = (0.2x)^2 - (0.8y)^2 = (0.2x - 0.8y)(0.2x + 0.8y)$$
Ответ:
- а) $$(x-6)(x+6)$$
- б) $$(1-a)(1+a)$$
- в) $$(4-x)(4+x)$$
- г) $$(9-y)(9+y)$$
- д) $$((\frac{1}{3}-b)(\frac{1}{3}+b)$$
- е) $$(y - \frac{5}{6})(y + \frac{5}{6})$$
- ж) $$(0.9 - x)(0.9 + x)$$
- з) $$(4a - 1)(4a + 1)$$
- и) $$(10 - 3y)(10 + 3y)$$
- к) $$(6a - 5b)(6a + 5b)$$
- л) $$(0.4q - 3p)(0.4q + 3p)$$
- м) $$((\frac{2}{9}k - \frac{1}{5}b)(\frac{2}{9}k + \frac{1}{5}b)$$
- н) $$((\frac{8}{3}n - 2m)(\frac{8}{3}n + 2m)$$
- о) $$(0.2x - 0.8y)(0.2x + 0.8y)$$