Question

сроооочно ,10 номер пожалуйста​

Answer 1

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