Bài 4 trang 189 SGK Đại số và Giải tích 12 Nâng cao

Đề bài

Thực hiện phép tính: \(\displaystyle {1 \over {2 - 3i}}\); \(\displaystyle {1 \over {{1 \over 2} - {{\sqrt 3 } \over 2}i}}\); \(\displaystyle {{3 - 2i} \over i}\); \(\displaystyle {{3 - 4i} \over {4 - i}}\)

Lời giải chi tiết

\(\displaystyle {1 \over {2 - 3i}} \) \(\displaystyle  = \frac{{2 + 3i}}{{\left( {2 - 3i} \right)\left( {2 + 3i} \right)}} \) \(\displaystyle  = {{2 + 3i} \over {4 - 9{i^2}}}  \) \(\displaystyle  = \frac{{2 + 3i}}{{13}}\) \(\displaystyle  = {2 \over {13}} + {3 \over {13}}i\)

\(\displaystyle {1 \over {{1 \over 2} - {{\sqrt 3 } \over 2}i}}  \) \(\displaystyle  = \frac{{\frac{1}{2} + \frac{{\sqrt 3 }}{2}i}}{{\left( {\frac{1}{2} - \frac{{\sqrt 3 }}{2}i} \right)\left( {\frac{1}{2} + \frac{{\sqrt 3 }}{2}i} \right)}}\) \(\displaystyle  = {{{1 \over 2} + {{\sqrt 3 } \over 2}i} \over {{1 \over 4} - {{\left( {{{\sqrt 3 } \over 2}i} \right)}^2}}}  \) \(\displaystyle  = {{{1 \over 2} + {{\sqrt 3 } \over 2}i} \over 1}  \) \(\displaystyle  = {1 \over 2} + {{\sqrt 3 } \over 2}i\)

\(\displaystyle {{3 - 2i} \over i}  \) \(\displaystyle  = {{i\left( {3 -  2i} \right)} \over {{i^2}}}  \) \(\displaystyle  = \frac{{3i - 2{i^2}}}{{ - 1}} \) \(\displaystyle =  - 3i + 2{i^2} \) \(\displaystyle =  - 3i - 2\)

\(\displaystyle {{3 - 4i} \over {4 - i}}  \) \(\displaystyle  = \frac{{\left( {3 - 4i} \right)\left( {4 + i} \right)}}{{\left( {4 - i} \right)\left( {4 + i} \right)}}\) \(\displaystyle = \frac{{12 - 13i - 4{i^2}}}{{{4^2} + {1^2}}} \) \(\displaystyle  = {{16 - 13i} \over {17}}  \) \(\displaystyle  = {{16} \over {17}} - {{13} \over {17}}i.\)

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