Упр.936 ГДЗ Никольский Потапов 9 класс (Алгебра)

б) a/(1-a)-(1-a^2)/(1+a^2)·(1/(a-2)^2-a/(1-a^2)) при a=2,5.

1. а) Подставим $$x=-1{,}5$$:

   $$ S=\left(\frac{4x}{4-x^2}-\frac{x-2}{4+2x}\right)\cdot \frac{4}{x+2}-\frac{x}{1-x^2} $$
   $$ S=\left(\frac{4\cdot(-1{,}5)}{4-(-1{,}5)^2}-\frac{-1{,}5-2}{4+2\cdot(-1{,}5)}\right)\cdot \frac{4}{-1{,}5+2}-\frac{-1{,}5}{1-(-1{,}5)^2} $$
   $$ S=\left(\frac{-6}{4-2{,}25}-\frac{-3{,}5}{4-3}\right)\cdot \frac{4}{0{,}5}+\frac{1{,}5}{1-2{,}25} $$
   $$ S=\left(\frac{-6}{1{,}75}-(-3{,}5)\right)\cdot 8-\frac{1{,}5}{1{,}25} $$
   $$ S=28-\frac{48}{1{,}75}-\frac{1{,}5}{1{,}25} $$
   $$ S=28-\frac{192}{7}-\frac{6}{5} $$
   $$ S=\frac{980-960-42}{35}=-\frac{22}{35} $$

2. б) Подставим $$a=2{,}5$$:

   $$ S=\frac{a}{1-a}-\frac{1-a^2}{1+a^2}\left(\frac{1}{(a-2)^2}-\frac{a}{1-a^2}\right) $$
   $$ S=\frac{2{,}5}{1-2{,}5}-\frac{1-2{,}5^2}{1+2{,}5^2}\left(\frac{1}{(2{,}5-2)^2}-\frac{2{,}5}{1-2{,}5^2}\right) $$
   $$ S=\frac{2{,}5}{-1{,}5}-\frac{1-6{,}25}{1+6{,}25}\left(\frac{1}{0{,}5^2}-\frac{2{,}5}{1-6{,}25}\right) $$
   $$ S=-\frac{5}{3}-\frac{-5{,}25}{7{,}25}\left(4+\frac{10}{21}\right) $$
   $$ S=-\frac{5}{3}+\frac{21}{29}\cdot \frac{94}{21} $$
   $$ S=-\frac{5}{3}+\frac{94}{29} $$
   $$ S=\frac{137}{87} $$

Ответ

а) $$-\frac{22}{35}$$; б) $$\frac{137}{87}$$.