Question

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Answer 1

Ответ:

$1)\ \ \displaystyle \frac{9c+27d}{9c}=\frac{9(c+3d)}{9c}=\frac{c+3d}{c}=1+\frac{3d}{c}\\\\\\2) \ \frac{6a-18b}{7a-21b}=\frac{6(a-3b)}{7(a-3b)}=\frac{6}{7}\\\\\\3)\ \ \frac{a^2-64}{3a+24}=\frac{(a-8)(a+8)}{3(a+8)}=\frac{a-8}{3}\\\\\\4)\ \ \frac{20x^2-5x}{6-24x}=\frac{5x(4x-1)}{-6(4x-1)}=-\frac{5x}{6}\\\\\\5)\ \ \frac{x^2-25}{x^2-10x+25}=\frac{(x-5)(x+5)}{(x-5)^2}=\frac{x+5}{x-5}$

$6)\ \ \dfrac{a^5-a^7}{a^6-a^4}=\dfrac{a^5(1-a^2)}{a^4(a^2-1)}=\dfrac{-a(a^2-1)}{a^2-1}=-a$