3) Đặt $z = a + bi\quad (a,\ b\in\Bbb R)$
$\Rightarrow \overline{z}= a - bi$
Ta được:
$\quad |\overline{z}|= |z + 2i|$
$\Leftrightarrow |a - bi|= |a + (b+2)i|$
$\Leftrightarrow a^2 + b^2 = a^2 + (b+2)^2$
$\Leftrightarrow 4b + 4= 0$
$\Leftrightarrow b = -1$
$\Rightarrow z = a - i$
Khi đó:
$\quad |z - i| + |z - 4|$
$= |a - 2i| + |a - 4 - i|$
$= \sqrt{a^2 + 4} + \sqrt{(a-4)^2 + 1}$
$\geqslant \sqrt{(a + 4 - a)^2 + (2+1)^2}\quad (BDT\ Minkowski)$
$= \sqrt{4^2 + 3^2}= 5$
Dấu $=$ xảy ra $\Leftrightarrow \dfrac a2 =\dfrac{4-a}{1}\Leftrightarrow a = \dfrac83$
Vậy $\min\left(|z - i| + |z - 4|\right)= 5 \Leftrightarrow z = \dfrac83 - i$
4) $I = \displaystyle\int\limits_{-\tfrac{\pi}{2}}^{\tfrac{\pi}{2}}f(x)dx$
$\to 2I = \displaystyle\int\limits_{-\tfrac{\pi}{2}}^{\tfrac{\pi}{2}}2f(x)dx$
$\to 2I = \displaystyle\int\limits_{-\tfrac{\pi}{2}}^{\tfrac{\pi}{2}}(\cos x - f(-x))dx$
$\to 2I = \displaystyle\int\limits_{-\tfrac{\pi}{2}}^{\tfrac{\pi}{2}}\cos xdx - \displaystyle\int\limits_{-\tfrac{\pi}{2}}^{\tfrac{\pi}{2}}f(-x)dx$
$\to 2I = \sin x\Bigg|_{-\tfrac{\pi}{2}}^{\tfrac{\pi}{2}} + \displaystyle\int\limits_{-\tfrac{\pi}{2}}^{\tfrac{\pi}{2}}f(-x)d(-x)$
$\to 2I = 2 + \displaystyle\int\limits_{-\tfrac{\pi}{2}}^{\tfrac{\pi}{2}}f(x)dx$
$\to 2I = 2 + I$
$\to I = 2$
