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Đáp án:
$\begin{array}{l}
a)\dfrac{1}{x}.\dfrac{{6x}}{y} = \dfrac{6}{y}\\
b)\dfrac{{2{x^2}}}{y}.3x{y^2} = 6{x^3}y\\
c)\dfrac{{15x}}{{7{y^3}}}.\dfrac{{2{y^2}}}{{{x^2}}} = \dfrac{{30}}{{7xy}}\\
d)\dfrac{{2{x^2}}}{{x - y}}.\dfrac{y}{{5{x^3}}} = \dfrac{{2y}}{{5x\left( {x - y} \right)}}\\
e)\dfrac{{5x + 10}}{{4x - 8}}.\dfrac{{4 - 2x}}{{x + 2}}\\
= \dfrac{{5\left( {x + 2} \right)}}{{4\left( {x - 2} \right)}}.\dfrac{{ - 2\left( {x - 2} \right)}}{{x + 2}}\\
= \dfrac{{ - 5}}{2}\\
f)\dfrac{{{x^2} - 36}}{{2x + 10}}.\dfrac{3}{{6 - x}}\\
= \dfrac{{\left( {x - 6} \right)\left( {x + 6} \right)}}{{2x + 10}}.\dfrac{3}{{6 - x}}\\
= \dfrac{{ - 3\left( {x + 6} \right)}}{{2x + 10}}\\
= - \dfrac{{3x + 18}}{{2x + 10}}\\
g)\dfrac{{{x^2} - 9{y^2}}}{{{x^2}{y^2}}}.\dfrac{{3xy}}{{2x - 6y}}\\
= \dfrac{{\left( {x - 3y} \right)\left( {x + 3y} \right)}}{{xy}}.\dfrac{3}{{2\left( {x - 3y} \right)}}\\
= \dfrac{{3\left( {x + 3y} \right)}}{{2xy}}\\
= \dfrac{{3x + 9y}}{{2xy}}\\
h)\dfrac{{3{x^2} - 3{y^2}}}{{5xy}}.\dfrac{{15{x^2}y}}{{2y - 2x}}\\
= \dfrac{{3\left( {x - y} \right)\left( {x + y} \right)}}{1}.\dfrac{{3x}}{{ - 2\left( {x - y} \right)}}\\
= - \dfrac{{9x\left( {x + y} \right)}}{2}\\
= - \dfrac{{9{x^2} + 9xy}}{2}\\
i)\dfrac{{2{a^3} - 2{b^3}}}{{3a + 3b}}.\dfrac{{6a + 6b}}{{{a^2} - 2ab + {b^2}}}\\
= \dfrac{{2\left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right).6\left( {a + b} \right)}}{{3\left( {a + b} \right).{{\left( {a - b} \right)}^2}}}\\
= \dfrac{{4\left( {{a^2} + ab + {b^2}} \right)}}{{a - b}}\\
= \dfrac{{4{a^2} + 4ab + 4{b^2}}}{{a - b}}
\end{array}$
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