Ответ:
а) \(\frac{16a^4}{21b^5}\cdot\frac{9b^2}{10a^3}=\frac{16\cdot9\cdot a^4b^2}{21\cdot10\cdot b^5a^3}=\frac{24a}{35b^3}\).
б) \(\frac{m^2n^3}{25t}\cdot\left(-\frac{5t}{mn^2}\right)=-\frac{m n}{5}\).
в) \(26mn^2\cdot\frac{3n^2}{4m^2}=\frac{78n^4}{4m}=\frac{39n^4}{2m}\).
г) \(\frac{2xy-y^2}{9}\cdot\frac{36}{y^4}=\frac{4y(2x-y)}{y^4}=\frac{4(2x-y)}{y^3}\).
д) \(\frac{a^2-7ab}{a^2+2ab}\cdot\frac{a^2+2ab}{a^2-7ab}=1\).
е) Разложим на множители: \(m^2-64=(m-8)(m+8)\), \(m^3-9m^2=m^2(m-9)\), \(m^2-81=(m-9)(m+9)\), \(m^2+8m=m(m+8)\). Поэтому \(\frac{m^2-64}{m^3-9m^2}\cdot\frac{m^2-81}{m^2+8m}=\frac{(m-8)(m+9)}{m^3}\).
ж) \(\frac{x^2-10x+25}{x^2-100}\colon\frac{x-5}{x-10}=\frac{(x-5)^2}{(x-10)(x+10)}\cdot\frac{x-10}{x-5}=\frac{x-5}{x+10}\).
з) \(\frac{a^2-1}{a-8}\colon\frac{a^2+2a+1}{a-8}=\frac{(a-1)(a+1)}{a-8}\cdot\frac{a-8}{(a+1)^2}=\frac{a-1}{a+1}\).
и) \(\frac{8a+b^2}{8b}\cdot\frac{ab+a^2}{2a}=\frac{8a+b^2}{8b}\cdot\frac{a(a+b)}{2a}=\frac{(8a+b^2)(a+b)}{16b}\).
к) \(\frac{x^2-16y^2}{25x^2-4y^2}\cdot\frac{x^2+8xy+16y^2}{25x^2+20xy+4y^2}=\frac{(x-4y)(x+4y)}{(5x-2y)(5x+2y)}\cdot\frac{(x+4y)^2}{(5x+2y)^2}=\frac{(x-4y)(x+4y)^3}{(5x-2y)(5x+2y)^3}\).
Ответ: а) \(\frac{24a}{35b^3}\); б) \(-\frac{mn}{5}\); в) \(\frac{39n^4}{2m}\); г) \(\frac{4(2x-y)}{y^3}\); д) \(1\); е) \(\frac{(m-8)(m+9)}{m^3}\); ж) \(\frac{x-5}{x+10}\); з) \(\frac{a-1}{a+1}\); и) \(\frac{(8a+b^2)(a+b)}{16b}\); к) \(\frac{(x-4y)(x+4y)^3}{(5x-2y)(5x+2y)^3}\).
