\(1)\ \frac{a^2-81}{a^2-8a}:\frac{a-9}{a^2-64}=\frac{a^2-81}{a^2-8a}\cdot\frac{a^2-64}{a-9}=\frac{ (a-9)(a+9)}{ a(a-8)}\cdot\frac{ (a-8)(a+8)}{a-9}=\frac{ (a-9)(a+9)\cdot(a-8)(a+8)}{ a(a-8)\cdot(a-9)}=\frac{ (a+9)\cdot(a+8)}{a\cdot1}=\frac{ (a+9)(a+8)}{a},\)
\(\phantom{1)\ }\begin{array}[t]{lcl}{\largeесли}\ a={-}4,\ {\largeто}&&\frac{ (a+9)(a+8)}{a}=\frac{ ({-}4+9)({-}4+8)}{{-}4}=\frac{5\cdot4}{{-}4}={-}\frac{20}{4}={-}5\end{array}\)
\(2)\ \frac{x}{4x^2-4y^2}:\frac{1}{6x+6y}=\frac{x}{4x^2-4y^2}\cdot\frac{6x+6y}{1}=\frac{x}{ 4(x^2-y^2)}\cdot\frac{ 6(x+y)}{1}=\frac{x}{ 4(x-y)(x+y)}\cdot\frac{ 6(x+y)}{1}=\frac{x\cdot6(x+y)}{ 4(x-y)(x+y)}=\frac{x\cdot3}{ 2(x-y)}=\frac{3x}{ 2(x-y)},\)
\(\phantom{2)\ }\begin{array}[t]{lcl}{\largeесли}\ x=4{,}2,\ y={-}2{,}8,\ {\largeто}&&\frac{3x}{ 2(x-y)}=\frac{3\cdot4{,}2}{ 2(4{,}2-({-}2{,}8))}=\frac{3\cdot4{,}2}{ 2(4{,}2+2{,}8)}=\frac{12{,}6}{2\cdot7}=\frac{12{,}6}{14}=\frac{126}{140}=\frac{9}{10}=0{,}9\end{array}\)
\(3)\ (3a^2-18a+27):\frac{3a-9}{4a}=\frac{3a^2-18a+27}{1}:\frac{3a-9}{4a}=\frac{3a^2-18a+27}{1}\cdot\frac{4a}{3a-9}=\frac{ 3(a^2-6a+9)}{1}\cdot\frac{4a}{ 3(a-3)}=\frac{ 3(a-3)^2}{1}\cdot\frac{4a}{ 3(a-3)}=\frac{ 3(a-3)^2\cdot4a}{ 3(a-3)}=\frac{ (a-3)\cdot4a}{1}=4a(a-3),\)
\(\phantom{3)\ }\begin{array}[t]{lcl}{\largeесли}\ a=0{,}5,\ {\largeто}&&4a(a-3)=4\cdot0{,}5(0{,}5-3)=2\cdot({-}2{,}5)={-}5\end{array}\)
\(4)\ \frac{a^6+a^5}{ (3a-3)^2}:\frac{a^5+a^4}{9a^2-9a}=\frac{a^6+a^5}{ (3a-3)^2}\cdot\frac{9a^2-9a}{a^5+a^4}=\frac{ a^5(a+1)}{ 9(a-1)^2}\cdot\frac{ 9a(a-1)}{ a^4(a+1)}=\frac{ a^5(a+1)\cdot9a(a-1)}{ 9(a-1)^2\cdot{a}^4(a+1)}=\frac{a\cdot{a}}{ (a-1)\cdot1}=\frac{a^2}{a-1},\)
\(\phantom{4)\ }\begin{array}[t]{lcl}{\largeесли}\ a=0{,}8,\ {\largeто}&&\frac{a^2}{a-1}=\frac{0{,}8^2}{0{,}8-1}=\frac{0{,}64}{{-}0{,}2}=\frac{64}{{-}20}={-}\frac{32}{10}={-}3{,}2\end{array}\)
