Question

Пж решите контрольку​

Answer 1

3)

$\frac{1}{x^{2}-4} -\frac{1}{x-2} -\frac{1}{x+2} =\frac{1}{(x-2)(x+2)} -\frac{1}{x-2} -\frac{1}{x+2}=\frac{1}{(x-2)(x+2)}-\frac{x+2}{(x-2)(x+2)} -\\\\-\frac{x-2}{(x-2)(x+2)} =\frac{1-x-2-x+2}{(x-2)(x+2)} =\frac{1-2x}{x^{2}-4}$

4)

$\frac{3a^{2}+6}{a^{3}+1} -\frac{3}{a^{2}-a+1} -\frac{1}{a+1} =\frac{3a^{2}+6}{(a+1)(a^{2}-a+1)} -\frac{3}{a^{2}-a+1} -\frac{1}{a+1}=\\\\=\frac{3a^{2}+6}{(a+1)(a^{2}-a+1)} -\frac{3(a+1)}{(a+1)(a^{2}-a+1)} -\frac{a^{2}-a+1}{(a+1)(a^{2}-a+1)} =\frac{3a^{2}+6-3a-3-a^{2}+a-1}{(a+1)(a^{2}-a+1)} =\frac{2a^{2}-2a+2}{(a+1)(a^{2}-a+1)} =\\\\=\frac{2(a^{2}-a+1)}{(a+1)(a^{2}-a+1)} =\frac{2}{a+1}$

$\frac{2}{a+1} =\frac{2}{-1.4+1} =\frac{2}{-0.4} =-5$