\(1)\ \frac{4-a}{8a^3}\cdot\frac{12a^5}{a^2-16}=\frac{ {-}(a-4)}{8a^3}\cdot\frac{12a^5}{a^2-16}={-}\frac{a-4}{8a^3}\cdot\frac{12a^5}{ (a-4)(a+4)}={-}\frac{ (a-4)\cdot12a^5}{8a^3\cdot(a-4)(a+4)}={-}\frac{1\cdot3a^2}{2\cdot(a+4)}={-}\frac{3a^2}{2a+8};\)
\(2)\ \frac{4c-d}{c^2+cd}\cdot\frac{2c^2-2d^2}{4c^2-cd}=\frac{4c-d}{ c(c+d)}\cdot\frac{ 2(c^2-d^2)}{ c(4c-d)}=\frac{4c-d}{ c(c+d)}\cdot\frac{ 2(c-d)(c+d)}{ c(4c-d)}=\frac{ (4c-d)\cdot2(c-d)(c+d)}{ c(c+d)\cdot{c}(4c-d)}=\frac{1\cdot2(c-d)}{c\cdot{c}}=\frac{2c-2d}{c^2};\)
\(3)\ \frac{b^2-6b+9}{b^2-3b+9}\cdot\frac{b^3+27}{5b-15}=\frac{ (b-3)^2}{b^2-3b+9}\cdot\frac{ (b+3)(b^2-3b+9)}{ 5(b-3)}=\frac{ (b-3)^2\cdot(b+3)(b^2-3b+9)}{ (b^2-3b+9)\cdot5(b-3)}=\frac{ (b-3)\cdot(b+3)}{1\cdot5}=\frac{b^2-9}{5};\)
\(4)\ \frac{a^3-16a}{3a^2b}\cdot\frac{12ab^2}{4a+16}=\frac{ a(a^2-16)}{3a^2b}\cdot\frac{12ab^2}{ 4(a+4)}=\frac{ a(a-4)(a+4)}{3a^2b}\cdot\frac{12ab^2}{ 4(a+4)}=\frac{ a(a-4)(a+4)\cdot12ab^2}{3a^2b\cdot4(a+4)}=\frac{ (a-4)\cdot{b}}{1\cdot1}=ab-4b;\)
\(5)\ \frac{a^3+b^3}{a^2-b^2}\cdot\frac{7a-7b}{a^2-ab+b^2}=\frac{ (a+b)(a^2-ab+b^2)}{ (a-b)(a+b)}\cdot\frac{ 7(a-b)}{a^2-ab+b^2}=\frac{ (a+b)(a^2-ab+b^2)\cdot7(a-b)}{ (a-b)(a+b)\cdot(a^2-ab+b^2)}=\frac{1\cdot7}{1\cdot1}=7;\)
\(6)\ \frac{x^2-9}{x+y}\cdot\frac{5x+5y}{x^2-3x}=\frac{ (x-3)(x+3)}{x+y}\cdot\frac{ 5(x+y)}{ x(x-3)}=\frac{ (x-3)(x+3)\cdot5(x+y)}{ (x+y)\cdot{x}(x-3)}=\frac{ (x+3)\cdot5}{1\cdot{x}}=\frac{5x+15}{x};\)
\(7)\ \frac{m+2n}{2-3m}:\frac{m^2+4mn+4n^2}{3m^2-2m}=\frac{m+2n}{2-3m}\cdot\frac{3m^2-2m}{m^2+4mn+4n^2}=\frac{m+2n}{ {-}(3m-2)}\cdot\frac{ m(3m-2)}{ (m+2n)^2}={-}\frac{ (m+2n)\cdot{m}(3m-2)}{ (3m-2)\cdot(m+2n)^2}={-}\frac{1\cdot{m}}{1\cdot(m+2n)}={-}\frac{m}{m+2n};\)
\(8)\ \frac{a^3+8}{16-a^4}:\frac{a^2-2a+4}{a^2+4}=\frac{a^3+8}{16-a^4}\cdot\frac{a^2+4}{a^2-2a+4}=\frac{ (a+2)(a^2-2a+4)}{ (4-a^2)(4+a^2)}\cdot\frac{4+a^2}{a^2-2a+4}=\frac{ (a+2)(a^2-2a+4)\cdot(4+a^2)}{ (4-a^2)(4+a^2)\cdot(a^2-2a+4)}=\frac{ (a+2)\cdot1}{ (4-a^2)\cdot1}=\frac{2+a}{ (2-a)(2+a)}=\frac{1}{2-a};\)
\(9)\ \frac{x^2-12x+36}{3x+21}\cdot\frac{x^2-49}{4x-24}=\frac{ (x-6)^2}{ 3(x+7)}\cdot\frac{ (x-7)(x+7)}{ 4(x-6)}=\frac{ (x-6)^2\cdot(x-7)(x+7)}{ 3(x+7)\cdot4(x-6)}=\frac{ (x-6)\cdot(x-7)}{3\cdot4}=\frac{x^2-7x-6x+42}{12}=\frac{x^2-13x+42}{12};\)
\(10)\ \frac{3a+15b}{a^2-81b^2}:\frac{4a+20b}{a^2-18ab+81b^2}=\frac{3a+15b}{a^2-81b^2}\cdot\frac{a^2-18ab+81b^2}{4a+20b}=\frac{ 3(a+5b)}{ (a-9b)(a+9b)}\cdot\frac{ (a-9b)^2}{ 4(a+5b)}=\frac{ 3(a+5b)\cdot(a-9b)^2}{ (a-9b)(a+9b)\cdot4(a+5b)}=\frac{3\cdot(a-9b)}{ (a+9b)\cdot4}=\frac{3a-27b}{4a+36b}.\)
