Упростите!1. [latex]( \frac{x+3}{x^2-3x}+ \frac{12}{9-x^2}) : \frac{3}{3+x} [/latex]2. [latex]( \frac{1}{d} -1) * \frac{4d}{1-d^2} [/latex]

Question

Упростите!1. [latex]( \frac{x+3}{x^2-3x}+ \frac{12}{9-x^2}) : \frac{3}{3+x} [/latex]2. [latex]( \frac{1}{d} -1) * \frac{4d}{1-d^2} [/latex]

Упростите! 1. [latex]( \frac{x+3}{x^2-3x}+ \frac{12}{9-x^2}) : \frac{3}{3+x} [/latex] 2. [latex]( \frac{1}{d} -1) * \frac{4d}{1-d^2} [/latex]

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Ответ(ы) на вопрос:

Accepted Answer

[latex]( \frac{x+3}{x^2-3x}+ \frac{12}{9-x^2}) : \frac{3}{3+x}= \frac{(x+3)^2-12x}{x(x-3)(x+3)}: \frac{3}{3+x}= \frac{(x-3)^2(3+x)}{3x(x-3)(x+3)}= \frac{x-3}{3x}[/latex] [latex]( \frac{1}{d} -1) \frac{4d}{1-d^2} = \frac{1-d}{d} \cdot\frac{4d}{1-d^2} = \frac{4d(1-d)}{d(1-d)(1+d)}= \frac{4}{1+d}[/latex]