Đáp án:
b) \(\dfrac{{x - 2}}{{2x - 1}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\dfrac{{{x^7} - {x^4}}}{{{x^6} - 1}} = \dfrac{{{x^4}\left( {{x^3} - 1} \right)}}{{{{\left( {{x^3}} \right)}^2} - 1}}\\
= \dfrac{{{x^4}\left( {{x^3} - 1} \right)}}{{\left( {{x^3} - 1} \right)\left( {{x^3} + 1} \right)}} = \dfrac{{{x^4}}}{{{x^3} + 1}}\\
b)\dfrac{{{x^4} - 2{x^3}}}{{2{x^4} - {x^3}}} = \dfrac{{{x^3}\left( {x - 2} \right)}}{{{x^3}\left( {2x - 1} \right)}}\\
= \dfrac{{x - 2}}{{2x - 1}}\\
c)\dfrac{{\left( {a - b} \right)\left( {c - d} \right)}}{{\left( {b - a} \right)\left( {b + a} \right)\left( {d - c} \right)\left( {d + c} \right)}}\\
= \dfrac{{\left( {a - b} \right)\left( {c - d} \right)}}{{\left( {a - b} \right)\left( {b + a} \right)\left( {c - d} \right)\left( {d + c} \right)}}\\
= \dfrac{1}{{\left( {a + b} \right)\left( {c + d} \right)}}
\end{array}\)
