Question

Алгебра 8 Рациональные уравнения (27)

Answer 1

$\displaystyle \tt \frac{2w-6}{w^2-64}-\frac{w-6}{w^2-8w}=\frac{w-1}{w^2+8w} \: \: \: \: | \: w\ne-8, \: w\ne8, \: w\ne0\\\frac{2w-6}{w^2-64}-\frac{w-6}{w^2-8w}-\frac{w-1}{w^2+8w}=0\\\frac{2w-6}{(w-8)(w+8)}-\frac{w-6}{w(w-8)}-\frac{w-1}{w(w+8)}=0\\\frac{w(2w-6)-(w+8)(w-6)-(w-8)(w-1)}{w(w-8)(w+8)}=0\\\frac{w(2w-6)-(w+8)(w-6)-(w^2-w-8w+8)}{w(w-8)(w+8)}=0\\\frac{2w^2-6w-w^2-2w+48-w^2+9w-8}{w(w-8)(w+8)}=0$

$\displaystyle \tt \frac{w+40}{w(w-8)(w+8)}=0\\w+40=0\\w=-40$