Đáp án:
7) \(\dfrac{1}{x}\)
Giải thích các bước giải:
\(\begin{array}{l}
7)\dfrac{{{x^2}\left( {2{x^3} - 128} \right)}}{{{x^3}\left( {x - 4} \right)\left( {2{x^2} + 8x + 32} \right)}}\\
= \dfrac{{2{x^2}\left( {{x^3} - 64} \right)}}{{2{x^3}\left( {x - 4} \right)\left( {{x^2} + 4x + 16} \right)}}\\
= \dfrac{{\left( {x - 4} \right)\left( {{x^2} + 4x + 16} \right)}}{{x\left( {x - 4} \right)\left( {{x^2} + 4x + 16} \right)}} = \dfrac{1}{x}\\
8)\dfrac{{{x^2} - 5x + 4}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} = \dfrac{{\left( {x - 4} \right)\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}\\
= \dfrac{{x - 4}}{{{x^2} + x + 1}}
\end{array}\)
