\(1)\ \frac{15a^2+10ab}{3ab+2b^2}=\frac{ 5a(3a+2b)}{ b(3a+2b)}=\frac{5a}{b}\)
\(\begin{array}[t]{lcl}{\largeЕсли}\ a={-}2,\ b=0{,}4,\ {\largeто}&&\frac{5a}{b}=\frac{ 5\cdot({-}2)}{0{,}4}=\frac{{-}10}{0{,}4}={-}\frac{100}{4}={-}25;\end{array}\)
\(2)\ \frac{9b^2-4c^2}{12b^2c-8bc^2}=\frac{ (3b-2c)(3b+2c)}{ 4bc(3b-2c)}=\frac{3b+2c}{4bc}\)
\(\begin{array}[t]{lcl}{\largeЕсли}\ b=\frac{1}{3},\ c={-}6,\ {\largeто}&&\frac{3b+2c}{4bc}=\frac{ 3\cdot\frac{1}{3}+2\cdot({-}6)}{ 4\cdot\frac{1}{3}\cdot({-}6)}=\frac{1-12}{ 4\cdot({-}2)}=\frac{{-}11}{{-}8}=\frac{11}{8}=1\frac{3}{8};\end{array}\)
\(3)\ \frac{36x^2-12xy+y^2}{y^2-36x^2}=\frac{ (6x-y)^2}{ (y-6x)(y+6x)}=\frac{ (y-6x)^2}{ (y-6x)(y+6x)}=\frac{y-6x}{y+6x}\)
\(\begin{array}[t]{lcl}{\largeЕсли}\ x=1{,}2,\ y={-}3,\ {\largeто}&&\frac{y-6x}{y+6x}=\frac{{-}3-6\cdot1{,}2}{{-}3+6\cdot1{,}2}=\frac{{-}3-7{,}2}{{-}3+7{,}2}=\frac{{-}10{,}2}{4{,}2}={-}\frac{102}{42}={-}\frac{17}{7}={-}2\frac{3}{7};\end{array}\)
\(4)\ \frac{a^8-a^6}{a^9+a^8}=\frac{ a^6(a^2-1)}{ a^8(a+1)}=\frac{ a^6(a-1)(a+1)}{ a^8(a+1)}=\frac{a-1}{a^2}\)
\(\begin{array}[t]{lcl}{\largeЕсли}\ a={-}0{,}1,\ {\largeто}&&\frac{a-1}{a^2}=\frac{{-}0{,}1-1}{ ({-}0{,}1)^2}=\frac{{-}1{,}1}{0{,}01}={-}110.\end{array}\)
