Question

Номер 19, под цифрой 2
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Answer 1

$\displaystyle \tt  \frac{1}{(x-1)(x-3)}+\frac{1}{(x-3)(x-5)}+\frac{1}{(x-5)(x-7)}=\frac{3}{(x-1)(x-7)};\\\\\\\\\frac{(x-5)(x-7)+(x-1)(x-7)+(x-1)(x-3)}{(x-1)(x-3)(x-5)(x-7)}=\\\\\\=\frac{x^{2}-5x-7x+35+x^{2}-x-7x+7+x^{2}-x-3x+3}{(x-1)(x-3)(x-5)(x-7)}=\\\\\\=\frac{3x^{2}-24x+45}{(x-1)(x-3)(x-5)(x-7)}=\frac{3\cdot(x^{2}-8x+15)}{(x-1)(x-3)(x-5)(x-7)}=\\\\\\=\frac{3\cdot(x-3)(x-5)}{(x-1)(x-3)(x-5)(x-7)}=\frac{3}{(x-1)(x-7)};$