Упр.932 ГДЗ Никольский Потапов 9 класс (Алгебра)
б) 1/(1-x)+1/(x+1)+2/(1+x^2)+4/(1+x^4)+8/(1+x^8)+16/(1+x^16).
а) Сгруппируем слагаемые попарно:
$$ \frac{1}{x(x+1)}+\frac{1}{(x+1)(x+2)}+\frac{1}{(x+2)(x+3)}+\frac{1}{(x+3)(x+4)}+\frac{1}{(x+4)(x+5)} $$
$$ =\frac{x+2+x}{x(x+1)(x+2)}+\frac{x+4+x+2}{(x+2)(x+3)(x+4)}+\frac{1}{(x+4)(x+5)} $$
$$ =\frac{2(x+1)}{x(x+1)(x+2)}+\frac{2(x+3)}{(x+2)(x+3)(x+4)}+\frac{1}{(x+4)(x+5)} $$
$$ =\frac{2}{x(x+2)}+\frac{2}{(x+2)(x+4)}+\frac{1}{(x+4)(x+5)} $$
$$ =\frac{2(x+4)+2x}{x(x+2)(x+4)}+\frac{1}{(x+4)(x+5)} =\frac{4(x+2)}{x(x+2)(x+4)}+\frac{1}{(x+4)(x+5)} $$
$$ =\frac{4}{x(x+4)}+\frac{1}{(x+4)(x+5)} =\frac{4(x+5)+x}{x(x+4)(x+5)} =\frac{5(x+4)}{x(x+4)(x+5)} $$
$$ =\frac{5}{x(x+5)}. $$
б) Последовательно используем формулу разности квадратов:
$$ \frac{1}{1-x}+\frac{1}{x+1}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}} $$
$$ =\frac{(x+1)+(1-x)}{(1-x)(1+x)}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}} $$
$$ =\frac{2}{1-x^2}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}} $$
$$ =\frac{2(1+x^2)+2(1-x^2)}{(1-x^2)(1+x^2)}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}} $$
$$ =\frac{4}{1-x^4}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}} $$
$$ =\frac{4(1+x^4)+4(1-x^4)}{(1-x^4)(1+x^4)}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}} $$
$$ =\frac{8}{1-x^8}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}} $$
$$ =\frac{8(1+x^8)+8(1-x^8)}{(1-x^8)(1+x^8)}+\frac{16}{1+x^{16}} =\frac{16}{1-x^{16}}+\frac{16}{1+x^{16}} $$
$$ =\frac{16(1+x^{16})+16(1-x^{16})}{(1-x^{16})(1+x^{16})} =\frac{32}{1-x^{32}}. $$
Ответ
а) $$\frac{5}{x(x+5)}$$; б) $$\frac{32}{1-x^{32}}$$.
