Đáp án A:
\(\dfrac{{{{\left( {1 + i} \right)}^5}}}{{{{\left( {1 - i} \right)}^3}}}\)\( = \dfrac{{{{\left( {1 + i} \right)}^4}.\left( {1 + i} \right)}}{{{{\left( {1 - i} \right)}^2}.\left( {1 - i} \right)}}\) \( = \dfrac{{{{\left( {2i} \right)}^2}\left( {1 + i} \right)}}{{\left( { - 2i} \right)\left( {1 - i} \right)}}\) \( = \dfrac{{ - 4\left( {1 + i} \right)}}{{ - 2\left( {1 + i} \right)}} = 2\) nên A sai.
Đáp án B:
\({\left( {1 + i} \right)^5} + {\left( {1 - i} \right)^5}\)\( = {\left( {1 + i} \right)^4}.\left( {1 + i} \right) + {\left( {1 - i} \right)^4}.\left( {1 - i} \right)\) \( = {\left( {2i} \right)^2}\left( {1 + i} \right) + {\left( { - 2i} \right)^2}\left( {1 - i} \right)\)
\( = - 4\left( {1 + i} \right) - 4\left( {1 - i} \right)\) \( = - 4 - 4i - 4 + 4i = - 8 \in \mathbb{R}\)
B sai.
Đáp án C:
\(\dfrac{{1 + i}}{{1 - i}} + \dfrac{{1 - i}}{{1 + i}}\)\( = \dfrac{{{{\left( {1 + i} \right)}^2} + {{\left( {1 - i} \right)}^2}}}{{\left( {1 - i} \right)\left( {1 + i} \right)}} = \dfrac{{2i - 2i}}{{1 + 1}} = 0\) nên C sai.
Đáp án D:
\(\dfrac{{3 + 2i}}{{2 - i}} - \dfrac{{3 - 2i}}{{2 + i}}\)\( = \dfrac{{\left( {3 + 2i} \right)\left( {2 + i} \right) - \left( {3 - 2i} \right)\left( {2 - i} \right)}}{{\left( {2 - i} \right)\left( {2 + i} \right)}}\) \( = \dfrac{{6 - 2 + 7i - 6 + 2 + 7i}}{{4 + 1}} = \dfrac{{14}}{5}i\)
Chọn D.
