Ответ:

1) а)

\(\left(\frac{x}{y^2}-\frac{1}{x}\right):\left(\frac{1}{y}-\frac{1}{x}\right)=\frac{x^2-y^2}{xy^2}:\frac{x-y}{xy}=\frac{(x-y)(x+y)}{xy^2}\cdot\frac{xy}{x-y}=\frac{x+y}{y}.\)

Ограничения: \(x\ne0\), \(y\ne0\), \(x\ne y\).

б)

\(\left(2+\frac{m}{m+1}\right)\cdot\frac{3m^2+3m}{12m+8}=\frac{3m+2}{m+1}\cdot\frac{3m(m+1)}{4(3m+2)}=\frac{3m}{4}.\)

Ограничения: \(m\ne-1\), \(m\ne-\frac{2}{3}\).

в)

\(\frac{4+b}{4-b}\cdot\left(\frac{2b^2}{4+b}-b\right)=\frac{4+b}{4-b}\cdot\frac{2b^2-b(4+b)}{4+b}=\frac{b(b-4)}{4-b}=-b.\)

Ограничения: \(b\ne-4\), \(b\ne4\).

г)

\(\left(\frac{y}{y-5}-2y\right):\frac{11-2y}{y-5}=\frac{y-2y(y-5)}{y-5}\cdot\frac{y-5}{11-2y}=\frac{y(11-2y)}{y-5}\cdot\frac{y-5}{11-2y}=y.\)

Ограничения: \(y\ne5\), \(y\ne\frac{11}{2}\).

д)

\(\frac{a+8b}{2b}-\frac{3a^2}{b^2}\cdot\frac{b}{6a}=\frac{a+8b}{2b}-\frac{a}{2b}=4.\)

Ограничения: \(a\ne0\), \(b\ne0\).

2) а)

\(\frac{x^2-4}{9-y^2}:\frac{x-2}{3+y}-\frac{2}{3-y}=\frac{(x-2)(x+2)}{(3-y)(3+y)}\cdot\frac{3+y}{x-2}-\frac{2}{3-y}=\frac{x+2}{3-y}-\frac{2}{3-y}=\frac{x}{3-y}.\)

Ограничения: \(x\ne2\), \(y\ne-3\), \(y\ne3\).

б)

\(\frac{a+b}{3a-b}+\frac{1}{a+b}\cdot\frac{a^2-b^2}{3a-b}=\frac{a+b}{3a-b}+\frac{a-b}{3a-b}=\frac{2a}{3a-b}.\)

Ограничения: \(3a-b\ne0\), \(a+b\ne0\).

в)

\(\left(\frac{1}{x-1}-\frac{x+1}{x^2+x+1}\right):\left(1+\frac{1}{x^3-1}\right)=\frac{x^2+x+1-(x+1)(x-1)}{(x-1)(x^2+x+1)}:\frac{x^3}{x^3-1}=\frac{x+2}{x^3-1}\cdot\frac{x^3-1}{x^3}=\frac{x+2}{x^3}.\)

Ограничения: \(x\ne1\), \(x\ne0\).