Câu 18 trang 214 SGK Giải tích 12 Nâng cao, Tính:...

Tính 

$\eqalign{ & a)\,\,{(\sqrt 3 + i)^2} - {(\sqrt 3 - i)^2} \cr & b)\,{(\sqrt 3 + i)^2} + {(\sqrt 3 - i)^2} \cr & c)\,{(\sqrt 3 + i)^3} - {(\sqrt 3 - i)^3} \cr & d)\,{{{{(\sqrt 3 + i)}^2}} \over {{{(\sqrt 3 - i)}^2}}} \cr}$

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) ${{{{(\sqrt 3  + i)}^2}} \over {{{(\sqrt 3  - i)}^2}}} = {{2 + 2\sqrt 3 i} \over {2 - 2\sqrt 3 i}} = {{1 + \sqrt 3 i} \over {1 - \sqrt 3 i}} = {{ - 1 + \sqrt 3 i} \over 2}$