Question

срочно пж решите пример
$\frac{y + 3}{2y + 2} - \frac{y + 1}{2y - 2} + \frac{3}{ {y}^{2} - 1}$
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Answer 1

$\frac{y+3}{2y+2}-\frac{y+1}{2y-2}+\frac{3}{y^2-1}=\frac{y+3}{2(y+1)}-\frac{y+1}{2(y-1)}+\frac{3}{(y-1)(y+1)}=\\\\=\frac{(y+3)(y-1)-(y+1)^2+3\cdot 2}{2(y-1)(y+1)}=\frac{y^2+2y-3-y^2-2y-1+6}{2(y-1)(y+1)}=\frac{2}{2(y-1)(y+1)}=\frac{1}{y^2-1}$

Answer 2

$\frac{y+3}{2y+2}-\frac{y+1}{2y-2}+\frac{3}{y^{2}-1}=\frac{y+3}{2(y+1)}-\frac{y+1}{2(y-1)}+\frac{3}{(y-1)(y+1)}=\frac{y^{2} -y+3y-3-y^{2}-2y-1+6 }{2(y-1)(y+1)}=\frac{2}{2(y^{2}-1)}=\frac{1}{y^{2} -1}$