\(1)\ \frac{2p}{5p-15}=\frac{2p}{ 5(p-3)}=\frac{ 2p(p^2+3p+9)}{ 5(p-3)(p^2+3p+9)}=\frac{ 2p(p^2+3p+9)}{ 5(p^3-27)}=\frac{2p^3+6p^2+18p}{5p^3-135}\)
\(\phantom{1)\ }\frac{1}{p^3-27}=\frac{1}{ (p-3)(p^2+3p+9)}=\frac{1\cdot5}{ (p-3)(p^2+3p+9)\cdot5}=\frac{5}{ 5(p^3-27)}=\frac{5}{5p^3-135};\)
\(2)\ \frac{3a+1}{9a^2-6a+1}=\frac{3a+1}{ (3a-1)^2}=\frac{ (3a+1)(3a+1)}{ (3a-1)^2(3a+1)}=\frac{ (3a+1)^2}{ (9a^2-6a+1)(3a+1)}=\frac{9a^2+6a+1}{27a^3-18a^2+3a+9a^2-6a+1}=\frac{9a^2+6a+1}{27a^3-9a^2-3a+1}\)
\(\phantom{2)\ }\frac{a-2}{9a^2-1}=\frac{a-2}{ (3a-1)(3a+1)}=\frac{ (a-2)(3a-1)}{ (3a-1)(3a+1)(3a-1)}=\frac{3a^2-a-6a+2}{ (3a-1)^2(3a+1)}=\frac{3a^2-7a+2}{ (9a^2-6a+1)(3a+1)}=\frac{3a^2-7a+2}{27a^3-18a^2+3a+9a^2-6a+1}=\frac{3a^2-7a+2}{27a^3-9a^2-3a+1};\)
\(3)\ \frac{a}{a^2-7a}=\frac{a}{ a(a-7)}=\frac{a-7}{ (a-7)(a-7)}=\frac{a-7}{ (a-7)^2}=\frac{a-7}{a^2-14a+49}\)
\(\phantom{3)\ }\frac{a+3}{a^2-14a+49}=\frac{a+3}{ (a-7)^2}=\frac{a+3}{a^2-14a+49};\)
\(4)\ \frac{2x}{x^2-1}=\frac{2x}{ (x-1)(x+1)}=\frac{ 2x(x-1)(x+1)}{ (x-1)(x+1)(x-1)(x+1)}=\frac{ 2x(x^2-1)}{ (x-1)^2(x+1)^2}=\frac{2x^3-2x}{ (x^2-1)^2}=\frac{2x^3-2x}{x^4-2x^2+1}\)
\(\phantom{4)\ }\frac{3x}{x^2-2x+1}=\frac{3x}{ (x-1)^2}=\frac{ 3x(x+1)^2}{ (x-1)^2(x+1)^2}=\frac{ 3x(x^2+2x+1)}{ (x^2-1)^2}=\frac{3x^3+6x^2+3x}{x^4-2x^2+1}\)
\(\phantom{4)\ }\frac{4}{x^2+2x+1}=\frac{4}{ (x+1)^2}=\frac{ 4(x-1)^2}{ (x+1)^2(x-1)^2}=\frac{ 4(x^2-2x+1)}{ (x^2-1)^2}=\frac{4x^2-8x+4}{x^4-2x^2+1};\)
\(5)\ \frac{a^2}{a^2-ab-ac+bc}=\frac{a^2}{ (a^2-ab)-(ac-bc)}=\frac{a^2}{ a(a-b)-c(a-b)}=\frac{a^2}{ (a-b)(a-c)}=\frac{4a^2}{ 4(a-b)(a-c)}=\frac{4a^2}{ 4(a^2-ab-ac+bc)}=\frac{4a^2}{4a^2-4ab-4ac+4bc}\)
\(\phantom{5)\ }\frac{b}{2a-2b}=\frac{b}{ 2(a-b)}=\frac{ 2b(a-c)}{ 2\cdot2(a-b)(a-c)}=\frac{2ab-2bc}{ 4(a^2-ab-ac+bc)}=\frac{2ab-2bc}{4a^2-4ab-4ac+4bc}\)
\(\phantom{5)\ }\frac{ab}{4a-4c}=\frac{ab}{ 4(a-c)}=\frac{ ab(a-b)}{ 4(a-c)(a-b)}=\frac{a^2b-ab^2}{ 4(a^2-ab-ac+bc)}=\frac{a^2b-ab^2}{4a^2-4ab-4ac+4bc}.\)