Đáp án:

\(\begin{array}{l}
a)x = {\left( {a + b} \right)^2}\\
b)x = \dfrac{{{{\left( {a + b} \right)}^2}}}{{a - b}}\\
c)x = \dfrac{1}{{{a^2} - {b^2}}}
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)x = \dfrac{{\left( {a - b} \right)\left( {a + b} \right)}}{{{a^2} + {b^2}}}:\dfrac{{{a^2} - 2ab + {b^2}}}{{{a^4} - {b^4}}}\\
 = \dfrac{{\left( {a - b} \right)\left( {a + b} \right)}}{{{a^2} + {b^2}}}.\dfrac{{\left( {{a^2} + {b^2}} \right)\left( {{a^2} - {b^2}} \right)}}{{{{\left( {a - b} \right)}^2}}}\\
 = \dfrac{{\left( {a + b} \right)\left( {{a^2} - {b^2}} \right)}}{{a - b}}\\
 = \dfrac{{\left( {a + b} \right)\left( {a - b} \right)\left( {a + b} \right)}}{{a - b}}\\
 = {\left( {a + b} \right)^2}\\
b)x = \dfrac{{\left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right)}}{{{{\left( {a - b} \right)}^2}}}:\dfrac{{{a^2} - ab + {b^2}}}{{\left( {a - b} \right)\left( {a + b} \right)}}\\
 = \dfrac{{\left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right)}}{{{{\left( {a - b} \right)}^2}}}.\dfrac{{\left( {a - b} \right)\left( {a + b} \right)}}{{{a^2} - ab + {b^2}}}\\
 = \dfrac{{{{\left( {a + b} \right)}^2}}}{{a - b}}\\
c)x = \dfrac{{a - b}}{{\left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right)}}:\dfrac{{{{\left( {a - b} \right)}^2}}}{{{a^2} - ab + {b^2}}}\\
 = \dfrac{{a - b}}{{\left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right)}}.\dfrac{{{a^2} - ab + {b^2}}}{{{{\left( {a - b} \right)}^2}}}\\
 = \dfrac{1}{{\left( {a + b} \right)\left( {a - b} \right)}}\\
 = \dfrac{1}{{{a^2} - {b^2}}}
\end{array}\)