Question

3, 4 срочно!!! даю 50 баллов

Answer 1

Ответ:

$\displaystyle 3)\ \ \Big(\dfrac{2x+5}{x^2+4x+4}-\frac{x+3}{x^2+2x}\Big):\frac{x^2-6}{x^3-4x}=\frac{x-2}{x+2}\\\\\\\Big(\dfrac{2x+5}{x^2+4x+4}-\frac{x+3}{x^2+2x}\Big):\frac{x^2-6}{x^3-4x}=\Big(\dfrac{2x+5}{(x+2)^2}-\frac{x+3}{x(x+2)}\Big):\frac{x^2-6}{x(x^2-4)}=\\\\\\=\dfrac{x(2x+5)-(x+3)(x+2)}{x(x+2)^2}\cdot \frac{x(x-2)(x+2)}{x^2-6}=$

$\displaystyle =\dfrac{2x^2+5x-x^2-5x-6}{x+2}\cdot \frac{x-2}{x^2-6}=\frac{x^2-6}{x+2}\cdot \frac{x-2}{x^2-6}=\frac{x-2}{x+2}$

$4)\ \ x^2+\dfrac{25}{x^2}=54\\\\\\\Big(x+\dfrac{5}{x}\Big)^2=x^2+10+\dfrac{25}{x^2}=\Big(x^2+\dfrac{25}{x^2}\Big)+10=54+10=64\ \ \Rightarrow \\\\\\x+\dfrac{5}{x}=\pm 8$