Giải thích các bước giải:

$\begin{aligned}
\text{1)} \quad & \dfrac{15x}{7y^3} \cdot \dfrac{2y^2}{x^2} \\
&= \dfrac{15x \cdot 2y^2}{7y^3 \cdot x^2} \\
&= \dfrac{30xy^2}{7x^2y^3} \\
&= \dfrac{30}{7xy} \\[10pt]
\text{2)} \quad & \dfrac{6x^3}{7y^4} \cdot \dfrac{35y^2}{24x} \\
&= \dfrac{6 \cdot 35 \cdot x^3 y^2}{7 \cdot 24 \cdot x y^4} \\
&= \dfrac{210x^3y^2}{168xy^4} \\
&= \dfrac{5x^2}{4y^2} \\[10pt]
\text{3)} \quad & \dfrac{4y^2}{x^4} \cdot \dfrac{-3x^2}{8y} \\
&= \dfrac{4y^2 \cdot (-3x^2)}{x^4 \cdot 8y} \\
&= \dfrac{-12x^2y^2}{8x^4y} \\
&= \dfrac{-3y}{2x^2} \\[10pt]
\text{4)} \quad & \dfrac{-18y^3}{25x^4} : \dfrac{9y^3}{-15x^2} \\
&= \dfrac{-18y^3}{25x^4} \cdot \dfrac{-15x^2}{9y^3} \\
&= \dfrac{270x^2y^3}{225x^4y^3} \\
&= \dfrac{6}{5x^2} \\[10pt]
\text{5)} \quad & \dfrac{8y^2}{9x^2} : \dfrac{4y}{3x^2} \\
&= \dfrac{8y^2}{9x^2} \cdot \dfrac{3x^2}{4y} \\
&= \dfrac{24x^2y^2}{36x^2y} \\
&= \dfrac{2y}{3} \\[10pt]
\text{6)} \quad & \dfrac{-20x}{3y^2} : \dfrac{-4x^3}{5y} \\
&= \dfrac{-20x}{3y^2} \cdot \dfrac{5y}{-4x^3} \\
&= \dfrac{-100xy}{-12x^3y^2} \\
&= \dfrac{25}{3x^2y} \\[10pt]
\text{7)} \quad & \dfrac{(x+4)^2}{4x+12} : \dfrac{x+4}{3x+9} \\
&= \dfrac{(x+4)^2}{4(x+3)} \cdot \dfrac{3(x+3)}{x+4} \\
&= \dfrac{3(x+4)^2(x+3)}{4(x+3)(x+4)} \\
&= \dfrac{3(x+4)}{4} \\[10pt]
\text{8)} \quad & \dfrac{5x+10}{4x-8} \cdot \dfrac{4-2x}{x+2} \\
&= \dfrac{5(x+2)}{4(x-2)} \cdot \dfrac{-2(x-2)}{x+2} \\
&= \dfrac{-10(x+2)(x-2)}{4(x-2)(x+2)} \\
&= \dfrac{-5}{2} \\[10pt]
\text{9)} \quad & \dfrac{x^2-36}{2x+10} \cdot \dfrac{3}{6-x} \\
&= \dfrac{(x-6)(x+6)}{2(x+5)} \cdot \dfrac{3}{-(x-6)} \\
&= \dfrac{3(x-6)(x+6)}{-2(x+5)(x-6)} \\
&= \dfrac{-3(x+6)}{2(x+5)} \\[10pt]
\text{10)} \quad & \dfrac{x^2-4}{x^2-x} : \dfrac{x^2+2x}{x-1} \\
&= \dfrac{(x-2)(x+2)}{x(x-1)} \cdot \dfrac{x-1}{x(x+2)} \\
&= \dfrac{(x-2)(x+2)(x-1)}{x^2(x-1)(x+2)} \\
&= \dfrac{x-2}{x^2} \\[10pt]
\text{11)} \quad & \dfrac{4(x+3)}{3x^2-x} : \dfrac{x^2+3x}{1-3x} \\
&= \dfrac{4(x+3)}{x(3x-1)} \cdot \dfrac{-(3x-1)}{x(x+3)} \\
&= \dfrac{-4(x+3)(3x-1)}{x^2(3x-1)(x+3)} \\
&= \dfrac{-4}{x^2} \\[10pt]
\text{12)} \quad & \dfrac{4x+12}{(x+4)^2} : \dfrac{3(x+3)}{x+4} \\
&= \dfrac{4(x+3)}{(x+4)^2} \cdot \dfrac{x+4}{3(x+3)} \\
&= \dfrac{4(x+3)(x+4)}{3(x+4)^2(x+3)} \\
&= \dfrac{4}{3(x+4)}
\end{aligned}$