Ответ:
- \[\frac{x-3}{3(x+2)}-\frac{x-6}{x+2}=\frac{x-3-3(x-6)}{3(x+2)}=\frac{15-2x}{3(x+2)}.\]
- \[\frac{m+4}{5(m-2)}+\frac{3-m}{4(m-2)}=\frac{4(m+4)+5(3-m)}{20(m-2)}=\frac{31-m}{20(m-2)}.\]
- \[\frac{(y+6)^2-(y+2)(y-6)}{(y-6)(y+6)}=\frac{16y+48}{y^2-36}=\frac{16(y+3)}{y^2-36}.\]
- Так как \(7-7x=-7(x-1)\), то
\[\frac{3x}{4(x-1)}-\frac{5x}{7(x-1)}=\frac{21x-20x}{28(x-1)}=\frac{x}{28(x-1)}.\] - \[\frac{2b}{2b+c}-\frac{4b^2}{(2b+c)^2}=\frac{2b(2b+c)-4b^2}{(2b+c)^2}=\frac{2bc}{(2b+c)^2}.\]
- \[\frac{2}{(a-3)(a+3)}-\frac{1}{a(a+3)}=\frac{2a-(a-3)}{a(a-3)(a+3)}=\frac{1}{a(a-3)}.\]
Ответ: \(\frac{15-2x}{3(x+2)};\ \frac{31-m}{20(m-2)};\ \frac{16(y+3)}{y^2-36};\ \frac{x}{28(x-1)};\ \frac{2bc}{(2b+c)^2};\ \frac{1}{a(a-3)}\).
