Đáp án:
e) \(\dfrac{{n - 1}}{{5\left( {n - 5} \right)}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\dfrac{{8x}}{{15{y^3}}}.\dfrac{{4{y^2}}}{{{x^2}}} = \dfrac{{32}}{{15xy}}\\
b)\dfrac{{9{a^2}}}{{a + 3}}.\dfrac{{\left( {a + 3} \right)\left( {a - 3} \right)}}{{6{a^3}}}\\
= \dfrac{{3\left( {a - 3} \right)}}{{2a}}\\
c)\dfrac{{4{n^2}}}{{17{m^4}}}.\left( { - \dfrac{{7{m^2}}}{{12n}}} \right)\\
= - \dfrac{{7n}}{{51{m^2}}}\\
d)\dfrac{{3\left( {b + 2} \right)}}{{{{\left( {b - 9} \right)}^2}}}.\dfrac{{2\left( {b - 9} \right)}}{{{{\left( {b + 2} \right)}^2}}}\\
= \dfrac{6}{{\left( {b - 9} \right)\left( {b + 2} \right)}}\\
e)\dfrac{{2{n^2} - 20n + 50}}{{5\left( {n + 1} \right)}}.\dfrac{{2\left( {n - 1} \right)\left( {n + 1} \right)}}{{4{{\left( {n - 5} \right)}^3}}}\\
= \dfrac{{2{{\left( {n - 5} \right)}^2}}}{{5\left( {n + 1} \right)}}.\dfrac{{2\left( {n - 1} \right)\left( {n + 1} \right)}}{{4{{\left( {n - 5} \right)}^3}}}\\
= \dfrac{{n - 1}}{{5\left( {n - 5} \right)}}
\end{array}\)
