10. Сократите дробь: a) \(\frac{2-2b^2}{4b^2-8b+4}\) б) \(\frac{3x^2-18x+27}{36-4x^2}\) в) \(\frac{a^3-3a^2+2a-6}{a^3-27}\)

Решение:

а)

\(\frac{2-2b^2}{4b^2-8b+4} = \frac{2(1-b^2)}{4(b^2-2b+1)} = \frac{2(1-b)(1+b)}{4(b-1)^2} = \frac{2(1-b)(1+b)}{-4(1-b)^2} = \frac{1+b}{-2(1-b)} = \frac{b+1}{2(b-1)}\)б)\(\frac{3x^2-18x+27}{36-4x^2} = \frac{3(x^2-6x+9)}{4(9-x^2)} = \frac{3(x-3)^2}{4(3-x)(3+x)} = \frac{3(x-3)^2}{-4(x-3)(3+x)} = \frac{3(x-3)}{-4(x+3)} = \frac{3(3-x)}{4(x+3)}\)в)\(\frac{a^3-3a^2+2a-6}{a^3-27} = \frac{a^2(a-3)+2(a-3)}{(a-3)(a^2+3a+9)} = \frac{(a^2+2)(a-3)}{(a-3)(a^2+3a+9)} = \frac{a^2+2}{a^2+3a+9}\)

Ответ: а) \(\frac{b+1}{2(b-1)}\), б) \(\frac{3(3-x)}{4(x+3)}\), в) \(\frac{a^2+2}{a^2+3a+9}\).