Question

помогите пожалуйста Сор​

Answer 1

Ответ:

а) 4(a+3)+a*(a-4) / a*(a+3)
4a+12+a^2-4a / a^2+3a
12+a^2 / a^2+3a
б) 2x^2 / (x-2)*(x+2) — 2x / x+2
2x^2-2x*(x-2) / (x-2)*(x+2)
2x^2-2x^2+4x / x^2-4
4x / x^2-4

Значит, а) 12+a^2 / a^2+3a
б) 4x / x^2-4

Answer 2

$\frac{4}{a} + \frac{a - 4}{a + 3} = \frac{4(a + 3) + a(a - 4)}{a(a + 3)} = \\ = \frac{4a + 12 + {a}^{2} - 4a}{ {a}^{2} + 3a} = \frac{ {a}^{2} + 12 }{ {a}^{2} + 3a}$

$\frac{2 {x}^{2} }{ {x}^{2} - 4} - \frac{2x}{x + 2} = \frac{2 {x}^{2} }{(x - 2)(x + 2)} - \frac{2x}{x + 2} = \\ = \frac{2 {x}^{2} - 2x(x - 2)}{(x - 2)(x + 2)} = \frac{2 {x}^{2} - 2 {x}^{2} + 4x}{ {x}^{2} - 4 } = \frac{4x}{ {x}^{2} - 4}$