Đáp án:
d) \( - \dfrac{5}{{a - 1}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\dfrac{{xy\left( {2{y^2} - {x^2}} \right)}}{{x\left( {x + y} \right)}} = \dfrac{{y\left( {2{y^2} - {x^2}} \right)}}{{x + y}}\\
b)\dfrac{{x\left( {{x^2} - 5x + 6} \right)}}{{ - 2\left( {2{x^2} - 5x + 2} \right)}} = \dfrac{{x\left( {x - 3} \right)\left( {x - 2} \right)}}{{ - 2\left( {x - 2} \right)\left( {2x - 1} \right)}}\\
= - \dfrac{{x\left( {x - 3} \right)}}{{2\left( {2x - 1} \right)}}\\
c)\dfrac{{{x^2} + 2xy + {y^2} - 16{y^2}}}{{3y\left( {x + 5y} \right)}} = \dfrac{{{{\left( {x + y} \right)}^2} - 16{y^2}}}{{3y\left( {x + 5y} \right)}}\\
= \dfrac{{\left( {x + y - 4y} \right)\left( {x + y + 4y} \right)}}{{3y\left( {x + 5y} \right)}}\\
= \dfrac{{\left( {x - 3y} \right)\left( {x + 5y} \right)}}{{3y\left( {x + 5y} \right)}}\\
= \dfrac{{x - 3y}}{{3y}}\\
d)\dfrac{{5\left( {a - 3} \right)}}{{ - \left( {{a^2} - 4a + 3} \right)}} = - \dfrac{{5\left( {a - 3} \right)}}{{\left( {a - 3} \right)\left( {a - 1} \right)}}\\
= - \dfrac{5}{{a - 1}}
\end{array}\)
