Bài tập 18 trang 214 SGK Toán 12 NC
Bài tập 18 trang 214 SGK Toán 12 NC
Tính
a) \({{{(\sqrt 3 + i)}^2} - {{(\sqrt 3 - i)}^2}}\)
b) \({{{(\sqrt 3 + i)}^2} + {{(\sqrt 3 - i)}^2}}\)
c) \({{{(\sqrt 3 + i)}^3} - {{(\sqrt 3 - i)}^3}}\)
d) \({\frac{{{{(\sqrt 3 + i)}^2}}}{{{{(\sqrt 3 - i)}^2}}}}\)
a)
\(\begin{array}{l}
{(\sqrt 3 + i)^2} - {(\sqrt 3 - i)^2}\\
= [\sqrt 3 + i + \sqrt 3 - i][\sqrt 3 + i - \sqrt 3 + i]\\
= 4\sqrt 3 i
\end{array}\)
b)
\(\begin{array}{l}
{(\sqrt 3 + i)^2} + {(\sqrt 3 - i)^2}\\
= 2 + 2\sqrt 3 i + 2 - 2\sqrt 3 i = 4
\end{array}\)
c)
\(\begin{array}{l}
{(\sqrt 3 + i)^2} - {(\sqrt 3 - i)^2}\\
= [\sqrt 3 + i - \sqrt 3 + i][{(\sqrt 3 + i)^2}\\
+ {(\sqrt 3 )^2} - {i^2} + {(\sqrt 3 - i)^2}]\\
= 2i(4 + 4) = 16i
\end{array}\)
d)
\(\begin{array}{l}
\frac{{{{(\sqrt 3 + i)}^2}}}{{{{(\sqrt 3 - i)}^2}}} = \frac{{2 + 2\sqrt 3 i}}{{2 - 2\sqrt 3 i}} = \frac{{1 + \sqrt 3 i}}{{1 - \sqrt 3 i}}\\
= \frac{{ - 1 + \sqrt 3 i}}{2}
\end{array}\)
-- Mod Toán 12
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