Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
a,\\
{x^2} + 1 - \dfrac{{{x^4} + 1}}{{{x^2} + 1}} = \dfrac{{{{\left( {{x^2} + 1} \right)}^2} - \left( {{x^4} + 1} \right)}}{{{x^2} + 1}} = \dfrac{{{x^4} + 2{x^2} + 1 - {x^4} - 1}}{{{x^2} + 1}} = \dfrac{{2{x^2}}}{{{x^2} + 1}}\\
b,\\
x + y - \dfrac{{{x^2} + {y^2}}}{{x + y}} = \dfrac{{{{\left( {x + y} \right)}^2} - \left( {{x^2} + {y^2}} \right)}}{{x + y}} = \dfrac{{\left( {{x^2} + 2xy + {y^2}} \right) - \left( {{x^2} + {y^2}} \right)}}{{x + y}} = \dfrac{{2xy}}{{x + y}}\\
c,\\
\dfrac{{1 + x}}{{x - 3}} - \dfrac{{1 - 2x}}{{3 + x}} - \dfrac{{x\left( {1 - x} \right)}}{{9 - {x^2}}}\\
 = \dfrac{{x + 1}}{{x - 3}} + \dfrac{{2x - 1}}{{x + 3}} + \dfrac{{x\left( {1 - x} \right)}}{{{x^2} - 9}}\\
 = \dfrac{{x + 1}}{{x - 3}} + \dfrac{{2x - 1}}{{x + 3}} + \dfrac{{x\left( {1 - x} \right)}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\
 = \dfrac{{\left( {x + 1} \right)\left( {x + 3} \right) + \left( {2x - 1} \right)\left( {x - 3} \right) + x\left( {1 - x} \right)}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\
 = \dfrac{{\left( {{x^2} + 4x + 3} \right) + \left( {2{x^2} - 7x + 3} \right) + x - {x^2}}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\
 = \dfrac{{2{x^2} - 2x + 6}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\
e,\\
\dfrac{1}{{a\left( {a - b} \right)\left( {a - c} \right)}} + \dfrac{1}{{b\left( {b - a} \right)\left( {b - c} \right)}} + \dfrac{1}{{c\left( {c - a} \right)\left( {c - b} \right)}}\\
 = \dfrac{1}{{a\left( {a - b} \right)\left( {a - c} \right)}} + \dfrac{1}{{b\left( {a - b} \right)\left( {c - b} \right)}} - \dfrac{1}{{c\left( {a - c} \right)\left( {c - b} \right)}}\\
 = \dfrac{{bc\left( {c - b} \right) + ac\left( {a - c} \right) - ab\left( {a - b} \right)}}{{abc\left( {a - b} \right)\left( {a - c} \right)\left( {c - b} \right)}}\\
 = \dfrac{{bc\left( {c - b} \right) + {a^2}c - a{c^2} - {a^2}b + a{b^2}}}{{abc\left( {a - b} \right)\left( {a - c} \right)\left( {c - b} \right)}}\\
 = \dfrac{{bc\left( {c - b} \right) + {a^2}\left( {c - b} \right) - a\left( {{c^2} - {b^2}} \right)}}{{abc\left( {a - b} \right)\left( {a - c} \right)\left( {c - b} \right)}}\\
 = \dfrac{{bc\left( {c - b} \right) + {a^2}\left( {c - b} \right) - a\left( {c - b} \right)\left( {c + b} \right)}}{{abc\left( {a - b} \right)\left( {a - c} \right)\left( {c - b} \right)}}\\
 = \dfrac{{\left( {c - b} \right).\left( {bc + {a^2} - ac - ab} \right)}}{{abc\left( {a - b} \right)\left( {a - c} \right)\left( {c - b} \right)}}\\
 = \dfrac{{\left( {c - b} \right)\left( {c\left( {b - a} \right) + a\left( {a - b} \right)} \right)}}{{abc\left( {a - b} \right)\left( {a - c} \right)\left( {c - b} \right)}}\\
 = \dfrac{{\left( {c - b} \right).\left( {a - b} \right)\left( {a - c} \right)}}{{abc\left( {a - b} \right)\left( {a - c} \right)\left( {c - b} \right)}}\\
 = \dfrac{1}{{abc}}
\end{array}\)