\({\largeа)\ }\begin{array}[t]{ll}{\largeПри}\ a={-}2,\ b={-}0{,}1\phantom{0,}&&\frac{15a^2-10ab}{3ab-2b^2}=\frac{ 5a(3a-2b)}{ b(3a-2b)}=\frac{5a}{b}=\frac{5\cdot({-}2)}{{-}0{,}1}=\frac{{-}10}{{-}0{,}1}=100;\end{array}\)
\({\largeб)\ }\begin{array}[t]{ll}{\largeПри}\ c=\frac{2}{3},\ d=\frac{1}{2}\phantom{0000,}&&\frac{9c^2-4d^2}{18c^2d-12cd^2}=\frac{ (3c-2d)(3c+2d)}{ 6cd(3c-2d)}=\frac{3c+2d}{6cd}=\frac{1}{2d}+\frac{1}{3c}=\frac{1}{2\cdot\frac{1}{2}}+\frac{1}{3\cdot\frac{2}{3}}=\frac{1}{1}+\frac{1}{2}=1\frac{1}{2}=1{,}5;\end{array}\)
\({\largeв)\ }\begin{array}[t]{ll}{\largeПри}\ x=\frac{2}{3},\ y={-}0{,}4\phantom{00}&&\frac{6x^2+12xy}{5xy+10y^2}=\frac{ 6x(x+2y)}{ 5y(x+2y)}=\frac{6x}{5y}=\frac{6\cdot\frac{2}{3}}{5\cdot({-}0{,}4)}=\frac{4}{{-}2}={-}2;\end{array}\)
\({\largeг)\ }\begin{array}[t]{ll}{\largeПри}\ x={-}0{,}2,\ y={-}0{,}6&&\frac{x^2+6xy+9y^2}{4x^2+12xy}=\frac{ (x+3y)^2}{ 4x(x+3y)}=\frac{x+3y}{4x}=\frac{{-}0{,}2+3\cdot({-}0{,}6)}{4\cdot({-}0{,}2)}=\frac{{-}0{,}2-1{,}8}{{-}0{,}8}=\frac{{-}2}{{-}0{,}8}=\frac{20}{8}=\frac{5}{2}=2\frac{1}{2}=2{,}5.\end{array}\)
