Đáp án:
$\begin{array}{l}
d)\dfrac{{{x^4}}}{{1 - x}} + {x^3} + {x^2} + x + 1\\
= \dfrac{{{x^4} + \left( {1 - x} \right)\left( {{x^3} + {x^2} + x + 1} \right)}}{{1 - x}}\\
= \dfrac{{{x^4} + {x^3} + {x^2} + x + 1 - {x^4} - {x^3} - {x^2} - 1}}{{1 - x}}\\
= \dfrac{x}{{1 - x}}\\
e)\dfrac{5}{{2{x^2}y}} + \dfrac{3}{{5x{y^2}}} + \dfrac{x}{{{y^3}}}\\
= \dfrac{{5.5x{y^2} + 3.2xy + x.10{x^2}}}{{10{x^2}{y^3}}}\\
= \dfrac{{25x{y^2} + 6xy + 10{x^3}}}{{10{x^2}{y^3}}}\\
f)\dfrac{{x + 1}}{{2x + 6}} + \dfrac{{2x + 3}}{{x\left( {x + 3} \right)}}\\
= \dfrac{{x + 1}}{{2\left( {x + 3} \right)}} + \dfrac{{2x + 3}}{{x\left( {x + 3} \right)}}\\
= \dfrac{{x\left( {x + 1} \right) + 2\left( {2x + 3} \right)}}{{2x\left( {x + 3} \right)}}\\
= \dfrac{{{x^2} + x + 4x + 6}}{{2x\left( {x + 3} \right)}}\\
= \dfrac{{{x^2} + 5x + 6}}{{2x\left( {x + 3} \right)}}\\
= \dfrac{{\left( {x + 2} \right)\left( {x + 3} \right)}}{{2x\left( {x + 3} \right)}}\\
= \dfrac{{x + 2}}{{2x}}\\
g){x^2} + \dfrac{{{x^4} + 1}}{{1 - {x^2}}} + 1\\
= \dfrac{{{x^2} - {x^4} + {x^4} + 1 + 1 - {x^2}}}{{1 - {x^2}}}\\
= \dfrac{2}{{1 - {x^2}}}
\end{array}$
