Question

Решить заранее ) a/a^2-1 + a^2+a+1/a^3-a^2+a-1 + a^2-a-1/a^3+a^2+a+1 - 2a^3/a^4-1

Answer 1

$\frac{a}{a^2-1}+ \frac{a^2+a+1}{a^3-a^2+a-1}+ \frac{a^2-a-1}{a^3+a^2+a+1}- \frac{2a^3}{a^4-1}= \\ = \frac{a}{a^2-1}+ \frac{a^2+a+1}{a^2(a-1)+1(a-1)}+ \frac{a^2-a-1}{a^2(a+1)+1(a+1)}- \frac{2a^3}{(a^2-1)(a^2+1)}= \\ =\frac{a}{(a-1)(a+1)}+ \frac{a^2+a+1}{(a^2+1)(a-1)}+ \frac{a^2-a-1}{(a^2+1)(a+1)}- \frac{2a^3}{(a-1)(a+1)(a^2+1)}= \\ = \frac{a^3+a+(a^2+a+1)(a+1)+(a^2-a-1)(a-1)-2a^3}{(a-1)(a+1)(a^2+1)}=\\=\frac{a^3+a+a^3+2a^2+2a+1+a^3-2a^2+1-2a^3}{(a-1)(a+1)(a^2+1)}= \\= \frac{a^3+3a+2}{(a-1)(a+1)(a^2+1)}$