Question

Упростите дробно-рациональное выражение 8p-40/15-3p 4-x2/10 - 5x (3x+6y)2/5x+10y ax+bx-ay-by/bx-by a2-6a+9/27-a3

Answer 1

$1) \frac{8p - 40}{15 - 3p} = \frac{8(p - 5)}{-3(-5 +p)} = \frac{8(p-5)}{-3(p-5)} = \frac{8}{-3} = - \frac{8}{3} = - 2 \frac{2}{3} \\ \\ \\ 2) \frac{4-x^2}{10-5x} = \frac{(2-x)(2+x)}{5(2-x)} = \frac{2+x}{5}= \frac{1}{5} (2+x)= 0.2(2+x)=0.4+0.2x \\ \\ \\ 3)\frac{(3x + 6y)^2}{5x+10y} = \frac{(3x+6y)(3x+6y)}{5(x+2y)} = \frac{3(x+2y) * 3(x+2y)}{5(x+2y)} = \\ \\ = \frac{9(x+2y)(x+2y)}{5(x+2y)} = \frac{9(x+2y)}{5} = 1.8(x+2y) =1.8x + 3.6y$
$4) \frac{ax+bx -ay-by}{bx-by} = \frac{a(x-y) +b(x-y))}{b(x-y)}= \frac{(a+b)(x-y)}{b(x-y)} = \\ \\ = \frac{a+b}{b} = \frac{a}{b}+ \frac{b}{b} = \frac{a}{b} + 1 \\ \\ 5) \frac{a^2-6a+9}{27-a^3} = \frac{a^2- 2*a*3 + 3^2}{-(a^3-3^3)} = \frac{(a-3)^2}{-(a-3)(a^2+3a+3^2)} = - \frac{a-3}{a^2+3a+9}$