Question

Помогите решить,СРОЧНО!МНОГО БАЛЛОВ)

Answer 1

$1) \frac{1}{1-a}-1= \frac{1-1+a}{1-a}= \frac{a}{1-a} \\\\a+ \frac{1-2 a^{2} }{1-a}+1= \frac{a- a^{2}+1-2 a^{2}+1-a }{1-a}= \frac{-3 a^{2}+2 }{1-a} \\\\ \frac{a}{1-a}: \frac{2-3 a^{2} }{1-a}= \frac{a}{1-a} * \frac{1-a}{2-3 a^{2} }= \frac{a}{2-3 a^{2} }$

$\frac{p}{ p^{2}-4 }+ \frac{2}{2-p} + \frac{1}{p+2}= \frac{p}{(p-2)(p+2)} - \frac{2}{p-2}+ \frac{1}{p+2} = \frac{p-2p-4+p-2}{(p-2)(p+2)}=$$=- \frac{6}{(p-2)(p+2)}\\\\\\p-2+ \frac{10- p^{2} }{p+2} = \frac{ p^{2}-4+10- p^{2} }{p+2}= \frac{6}{p+2} \\\\\\- \frac{6}{(p-2)(p+2)} : \frac{6}{p+2} =- \frac{6}{(p+2)(p-2)} * \frac{p+2}{6}= \frac{1}{2-p}$