Ответ:
1) \(3x+6=3(x+2)\):
\[\frac{x-3}{3(x+2)}-\frac{x-6}{x+2}=\frac{x-3-3(x-6)}{3(x+2)}=\frac{-2x+15}{3(x+2)}.\]
ОДЗ: \(x\ne-2\).
2) Разложим знаменатели на множители:
\[5m-40=5(m-8),\qquad 4m-8=4(m-2).\]
\[\frac{m+4}{5(m-8)}+\frac{3-m}{4(m-2)}=\frac{4(m+4)(m-2)+5(3-m)(m-8)}{20(m-8)(m-2)}=\frac{-m^2+63m-152}{20(m-8)(m-2)}.\]
ОДЗ: \(m\ne8\), \(m\ne2\).
3) Общий знаменатель: \((y-6)(y+6)=y^2-36\).
\[\frac{y+6}{y-6}-\frac{y+2}{y+6}=\frac{(y+6)^2-(y+2)(y-6)}{y^2-36}=\frac{16y+48}{y^2-36}=\frac{16(y+3)}{y^2-36}.\]
ОДЗ: \(y\ne6\), \(y\ne-6\).
4) Учтём, что \(7-7x=-7(x-1)\), а \(4x-4=4(x-1)\):
\[\frac{3x}{4(x-1)}+\frac{5x}{-7(x-1)}=\frac{21x-20x}{28(x-1)}=\frac{x}{28(x-1)}.\]
ОДЗ: \(x\ne1\).
5) \(4b^2+4bc+c^2=(2b+c)^2\):
\[\frac{2b}{2b+c}-\frac{4b^2}{(2b+c)^2}=\frac{2b(2b+c)-4b^2}{(2b+c)^2}=\frac{2bc}{(2b+c)^2}.\]
ОДЗ: \(2b+c\ne0\).
6) \(a^2-9=(a-3)(a+3)\), \(a^2+3a=a(a+3)\):
\[\frac{2}{(a-3)(a+3)}-\frac{1}{a(a+3)}=\frac{2a-(a-3)}{a(a-3)(a+3)}=\frac{a+3}{a(a-3)(a+3)}=\frac{1}{a(a-3)}.\]
ОДЗ: \(a\ne-3\), \(a\ne0\), \(a\ne3\).
Ответ: 1) \(\frac{-2x+15}{3(x+2)}\); 2) \(\frac{-m^2+63m-152}{20(m-8)(m-2)}\); 3) \(\frac{16(y+3)}{y^2-36}\); 4) \(\frac{x}{28(x-1)}\); 5) \(\frac{2bc}{(2b+c)^2}\); 6) \(\frac{1}{a(a-3)}\).
