Question

ОЧЕНЬ ОЧЕНЬ СРОЧНО ПОМОЖІТЬ ПЖ​

Answer 1

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

$= \frac{a+b+a-b}{a^2-b^2} +\frac{2a}{a^2+b^2} +\frac{4a^3}{a^4+b^4} =$

$= \frac{2a}{a^2-b^2} +\frac{2a}{a^2+b^2} +\frac{4a^3}{a^4+b^4} =$

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

$= \frac{2a^3+2ab^2}{a^4-b^4} +\frac{2a^3-2ab^2}{a^4-b^4} +\frac{4a^3}{a^4+b^4} =$

$= \frac{4a^3}{a^4-b^4} +\frac{4a^3}{a^4+b^4} =$

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

$= \frac{4a^7+4a^3b^4}{a^8-b^8} +\frac{{4a^7-4a^3b^4}}{a^8-b^8}} =$

$= \frac{4a^7}{a^8-b^8}$