Nội dung bài giảng
Tính
\(\eqalign{
& a)\,\,{(\sqrt 3 + i)^2} - {(\sqrt 3 - i)^2} \cr
& b)\,{(\sqrt 3 + i)^2} + {(\sqrt 3 - i)^2} \cr
& c)\,{(\sqrt 3 + i)^3} - {(\sqrt 3 - i)^3} \cr
& d)\,{{{{(\sqrt 3 + i)}^2}} \over {{{(\sqrt 3 - i)}^2}}} \cr} \)
Giải
a)
\(\eqalign{
& {(\sqrt 3 + i)^2} - {(\sqrt 3 - i)^2} \cr&= {\rm{[}}\sqrt 3 + i + \sqrt 3 - i{\rm{][}}\sqrt 3 + i - \sqrt 3 + i{\rm{]}} \cr
& {\rm{ = 4}}\sqrt 3 i \cr} \)
b)
\({(\sqrt 3 + i)^2} + {(\sqrt 3 - i)^2} = 2 + 2\sqrt 3 i + 2 - 2\sqrt 3 i = 4\)
c)
\(\eqalign{
& {(\sqrt 3 + i)^2} - {(\sqrt 3 - i)^2} = {\rm{[}}\sqrt 3 + i - \sqrt 3 + i{\rm{][}}{(\sqrt 3 + i)^2} + {(\sqrt 3 )^2} - {i^2} + {(\sqrt 3 - i)^2}{\rm{]}} \cr
& = 2i(4 + 4) = 16i \cr} \)
d) \({{{{(\sqrt 3 + i)}^2}} \over {{{(\sqrt 3 - i)}^2}}} = {{2 + 2\sqrt 3 i} \over {2 - 2\sqrt 3 i}} = {{1 + \sqrt 3 i} \over {1 - \sqrt 3 i}} = {{ - 1 + \sqrt 3 i} \over 2}\)