Question

Доведите Тотожность, на фото 17.31

Answer 1

$\frac{2}{x^2-1}+\frac{4}{x^2-4}+\frac{6}{x^2-9}+\frac{8}{x^2-16}=\\\\=\frac{2}{(x-1)(x+1)}+\frac{8}{(x-4)(x+4)}+\frac{4}{(x-2)(x+2)}+\frac{6}{(x-3)(x+3)}=\\\\=\frac{2(x-4)(x+4)+8(x-1)(x+1)}{(x-1)(x+1)(x-4)(x+4)}+\frac{4(x-3)(x+3)+6(x-2)(x+2)}{(x-2)(x+2)(x-3)(x+3)}=\\\\=\frac{2x^2-32+8x^2-8}{(x-1)(x+1)(x-4)(x+4)}+\frac{4x^2-36+6x^2-24}{(x-2)(x+2)(x-3)(x+3)}=\\\\=\frac{10x^2-40}{(x-1)(x+1)(x-4)(x+4)}+\frac{10x^2-60}{(x-2)(x+2)(x-3)(x+3)}=$

$\\=\frac{5x^2-15x-20+5x^2+15x-20}{(x-4)(x+1)(x-1)(x+4)}+\frac{5x^2-5x-30+5x^2+5x-30}{(x-2)(x+3)(x-3)(x+2)}=\\\\=\frac{5(x-4)(x+1)+5(x-1)(x+4)}{(x-4)(x+1)(x-1)(x+4)}+\frac{5(x-3)(x+2)+5(x-2)(x+3)}{(x-2)(x+3)(x-3)(x+2)}=\\\\=\frac{5}{(x-1)(x+4)}+\frac{5}{(x-4)(x+1)}+\frac{5}{(x-2)(x+3)}+\frac{5}{(x-3)(x+2)}=\\\\=5(\frac{1}{(x-1)(x+4)}+\frac{1}{(x-2)(x+3)}+\frac{1}{(x-3)(x+2)}+\frac{1}{(x-4)(x+1)})$