Question

помогите решить задание.

Answer 1

№1

а) $\frac{14x^{4}y}{49x^{3}y^{2}} =  \frac{2x}{7y}$

б) $\frac{5x}{x^{2} + 3x}  = \frac{5}{x + 3}$

в) не знаю

г) $\frac{x^{2} - 16}{x^{2} - 8x + 16} = \frac{-16}{-8y+16}$

Answer 2

1. a)$a) \frac{14x^4y}{49x^3y^2} = \frac{2x * 7x^3y}{7y*7x^3y} = \frac{2x}{7y}\\b) \frac{5x}{x^2 + 3x} = \frac{5x}{x(x + 3)} = \frac{5}{x+3}\\c) \frac{x^2 - y^2}{2x - 2y} = \frac{(x-y)(x+y)}{2(x-y)} = \frac{x+y}{2}\\d) \frac{x^2-16}{x^2-8x+16} = \frac{(x-4)(x+4)}{(x-4)^2} = \frac{x+4}{x-4}$

2. $a) \frac{3x - 1}{x^2} + \frac{x-9}{3x} = \frac{9x - 3 + x^2 - 9x}{3x^2} = \frac{x^2 - 3}{3x^2}\\b) \frac{1}{2a-b} - \frac{1}{2a+b} = \frac{2a+b-2a+b}{(2a-b)(2a+b)} = \frac{2b}{(2a-b)(2a+b)}\\c) \frac{5}{c+3} - \frac{5c-2}{c^2 + 3c} = \frac{5c - 5c + 2}{c(c+3)} = \frac{2}{c(c+3)}$

3. $\frac{2a}{a-5} - \frac{5}{a+5} + \frac{2a^2}{25-a^2} = \frac{2a^2 + 10a - 5a + 25 + 2a^2}{(a-5)(a+5)} = \frac{4a^2 + 5a + 25}{(a-5)(a+5)}$

4.$\frac{2x^2 + 7x + 9}{x^3 - 1} + \frac{4x+3}{x^2+x+1} - \frac{5}{x-1} = \frac{2x^2 + 7x + 9+4x^2-x-3-5x^2-5x-5}{(x-1)(x^2 + x + 1)} = \frac{x^2 + x+1}{(x-1)(x^2 + x+1)} = \frac{1}{x-1} = \frac{1}{0.1} = 10$