Question

Решите СРОЧНО
$\sqrt{2x - 4} < 4$
$\sqrt{5x + 1} \geqslant \sqrt{6x - 3 }$
Найти сумму, разность и деление.
1) z1=1+2i
2) z2=4-3i

Answer 1

Смотри......................

Answer 2

$1)\; \; \; \; \sqrt{2x-4}<4\; \; ,\; \; \; ODZ:\; 2x-4\geq 0\; ,\; \; x\geq 2\\\\2x-4<16\; ,\; \; 2x<20\; ,\; \; x<10\\\\x\in [\, 2,10)\\\\2)\; \; \sqrt{5x+1}\geq \sqrt{6x-3}\; ,\\\\\sqrt{5x+1}\geq \sqrt{6x-3}\geq 0\; \; ,\; \; ODZ:\; 6x-3\geq 0\; ,\; \; x\geq \frac{1}{2}\\\\5x+1\geq 6x-3\\\\1+3\geq 6x-5x\\\\4\geq x\\\\x\leq 4\\\\x\in [\; \frac{1}{2},4\; ]$

$3)\; \; z_1=1+2i\; \; ,\; \; z_2=4-3i\\\\z_1+z_2=(1+2i)+(4-3i)=5-i\\\\z_1-z_2=(1+2i)-(4-3i)=-3+5i\\\\\frac{z_1}{z_2}=\frac{(1+2i)(4+3i)}{(4-3i)(4+3i)}=\frac{4+3i+8i+6i^2}{4^2-9i^2}=\frac{4+11i-6}{16+9}=\frac{-2+11i}{25}=-\frac{2}{25}+\frac{11}{25}\, i$