## 30 номер

### Пример:

$x^3+1=0$

### Решение:

$x^3+1=(x+1)(x^2-x+1)$
$x = -1;$
$x^2-x+1=0$
$D = -3$

$\text{Ответ: } x=\dfrac{1\pm i\sqrt{ 3 }}{2}$

## 31 номер

### Пример:

$x^4-4x^2+5=0$

### Решение:

$t=x^{2}$
$t^2-4t+5=0$
$D=-4$
$t=\dfrac{4\pm 2i}{2}=2\pm i$

$x^2=2\pm i$

$\text{1. }x^2=2+i$
$x=a+bi$
$(a+bi)^2=(a^2-b^2)+2abi=2+i$

$$
\begin{cases}
a^2-b^2=2 \\
2abi=i; 2ab=1
\end{cases}
$$

$(a^2+b^2)^2=(a^2-b^2)^2+(2ab)^2=2^2+1^2=5$
$a^2+b^2=\sqrt{5}$
$a^2=\dfrac{(a^2+b^2)+(a^2-b^2)}{2}=\dfrac{\sqrt{ 5 }+2}{2}$
$b^2=\dfrac{(a^2+b^2)-(a^2-b^2)}{2}=\dfrac{\sqrt{ 5 }-2}{2}$
$x_{1}=\sqrt{ \dfrac{\sqrt{ 5 }+2}{2} }+i\sqrt{ \dfrac{\sqrt{ 5 }-2}{2} }$
$x_{2}=-( \sqrt{ \dfrac{\sqrt{ 5 }+2}{2} }+i\sqrt{ \dfrac{\sqrt{ 5 }-2}{2} } )$

$\text{2. }x^2=2-i$
$2-i=\overline{(2+i)}$
$x_{3}=\sqrt{ \dfrac{\sqrt{ 5 }+2}{2} }-i\sqrt{ \dfrac{\sqrt{ 5 }-2}{2} }$
$x_{4}=-( \sqrt{ \dfrac{\sqrt{ 5 }+2}{2} }-i\sqrt{ \dfrac{\sqrt{ 5 }-2}{2} } )$

$\text{Ответ:}$

$$
\begin{array} \\
x_{1}=\sqrt{ \dfrac{\sqrt{ 5 }+2}{2} }+i\sqrt{ \dfrac{\sqrt{ 5 }-2}{2} } \\
x_{2}=-( \sqrt{ \dfrac{\sqrt{ 5 }+2}{2} }+i\sqrt{ \dfrac{\sqrt{ 5 }-2}{2} } ) \\
x_{3}=\sqrt{ \dfrac{\sqrt{ 5 }+2}{2} }-i\sqrt{ \dfrac{\sqrt{ 5 }-2}{2} } \\
x_{4}=-( \sqrt{ \dfrac{\sqrt{ 5 }+2}{2} }-i\sqrt{ \dfrac{\sqrt{ 5 }-2}{2} } )
\end{array}
$$

## 32 номер

### Пример:

$x^4+4x^2+20=0$

### Решение:

$t=x^{2}$
$t^2+4t+20=0$
$D=-64$
$t=\dfrac{-4\pm 8i}{2}=-2\pm 4i$

$x^2=-2\pm 4i$

$\text{1. }x^2=-2+4i$
$x=a+bi$
$(a+bi)^2=(a^2-b^2)+2abi=-2+4i$

$$
\begin{cases}
a^2-b^2=-2 \\
2ab=4
\end{cases}
$$

$(a^2+b^2)^2=(a^2-b^2)^2+(2ab)^2=(-2)^2+4^2=20$
$a^2+b^2=\sqrt{20}=2\sqrt5$
$a^2=\dfrac{(a^2+b^2)+(a^2-b^2)}{2}=\dfrac{2\sqrt5-2}{2}=\sqrt5-1$
$b^2=\dfrac{(a^2+b^2)-(a^2-b^2)}{2}=\dfrac{2\sqrt5+2}{2}=\sqrt5+1$

$x_{1}=\sqrt{\sqrt5-1}+i\sqrt{\sqrt5+1}$
$x_{2}=-(\sqrt{\sqrt5-1}+i\sqrt{\sqrt5+1})$

$\text{2. }x^2=-2-4i$
$-2-4i=\overline{(-2+4i)}$

$x_{3}=\sqrt{\sqrt5-1}-i\sqrt{\sqrt5+1}$
$x_{4}=-(\sqrt{\sqrt5-1}-i\sqrt{\sqrt5+1})$

$\text{Ответ:}$

$$
\begin{array}{l}
x_{1}=\sqrt{\sqrt5-1}+i\sqrt{\sqrt5+1}\\
x_{2}=-(\sqrt{\sqrt5-1}+i\sqrt{\sqrt5+1})\\
x_{3}=\sqrt{\sqrt5-1}-i\sqrt{\sqrt5+1}\\
x_{4}=-(\sqrt{\sqrt5-1}-i\sqrt{\sqrt5+1})
\end{array}
$$

## 33 номер

### Пример:

$x^4-6x^2+13=0$

### Решение:

$t=x^{2}$
$t^2-6t+13=0$
$D=-16$
$t=\dfrac{6\pm4i}{2}=3\pm2i$

$x^2=3\pm2i$

$\text{1. }x^2=3+2i$
$x=a+bi$
$(a+bi)^2=(a^2-b^2)+2abi=3+2i$

$$
\begin{cases}
a^2-b^2=3 \\
2ab=2
\end{cases}
$$

$(a^2+b^2)^2=(a^2-b^2)^2+(2ab)^2=3^2+2^2=13$
$a^2+b^2=\sqrt{13}$
$a^2=\dfrac{(a^2+b^2)+(a^2-b^2)}{2}=\dfrac{\sqrt{13}+3}{2}$
$b^2=\dfrac{(a^2+b^2)-(a^2-b^2)}{2}=\dfrac{\sqrt{13}-3}{2}$

$x_{1}=\sqrt{\dfrac{\sqrt{13}+3}{2}}+i\sqrt{\dfrac{\sqrt{13}-3}{2}}$
$x_{2}=-(\sqrt{\dfrac{\sqrt{13}+3}{2}}+i\sqrt{\dfrac{\sqrt{13}-3}{2}})$

$\text{2. }x^2=3-2i$
$3-2i=\overline{(3+2i)}$

$x_{3}=\sqrt{\dfrac{\sqrt{13}+3}{2}}-i\sqrt{\dfrac{\sqrt{13}-3}{2}}$
$x_{4}=-(\sqrt{\dfrac{\sqrt{13}+3}{2}}-i\sqrt{\dfrac{\sqrt{13}-3}{2}})$

$\text{Ответ:}$

$$
\begin{array}{l}
x_{1}=\sqrt{\dfrac{\sqrt{13}+3}{2}}+i\sqrt{\dfrac{\sqrt{13}-3}{2}}\\
x_{2}=-(\sqrt{\dfrac{\sqrt{13}+3}{2}}+i\sqrt{\dfrac{\sqrt{13}-3}{2}})\\
x_{3}=\sqrt{\dfrac{\sqrt{13}+3}{2}}-i\sqrt{\dfrac{\sqrt{13}-3}{2}}\\
x_{4}=-(\sqrt{\dfrac{\sqrt{13}+3}{2}}-i\sqrt{\dfrac{\sqrt{13}-3}{2}})
\end{array}
$$

## 34 номер

### Пример:

$x^4 + 2x^2 + 17 = 0$

### Решение:

$t=x^{2}$
$t^2+2t+17=0$
$D=-64$
$t=\dfrac{-2\pm 8i}{2}=-1\pm 4i$

$x^2=-1\pm 4i$

$\text{1. }x^2=-1+4i$
$x=a+bi$
$(a+bi)^2=(a^2-b^2)+2abi=-1+4i$

$$
\begin{cases}
a^2-b^2=-1 \\
2ab=4
\end{cases}
$$

$(a^2+b^2)^2=(a^2-b^2)^2+(2ab)^2=(-1)^2+4^2=17$
$a^2+b^2=\sqrt{17}$
$a^2=\dfrac{(a^2+b^2)+(a^2-b^2)}{2}=\dfrac{\sqrt{17}-1}{2}$
$b^2=\dfrac{(a^2+b^2)-(a^2-b^2)}{2}=\dfrac{\sqrt{17}+1}{2}$

$x_{1}=\sqrt{\dfrac{\sqrt{17}-1}{2}}+i\sqrt{\dfrac{\sqrt{17}+1}{2}}$
$x_{2}=-(\sqrt{\dfrac{\sqrt{17}-1}{2}}+i\sqrt{\dfrac{\sqrt{17}+1}{2}})$

$\text{2. }x^2=-1-4i$
$-1-4i=\overline{(-1+4i)}$

$x_{3}=\sqrt{\dfrac{\sqrt{17}-1}{2}}-i\sqrt{\dfrac{\sqrt{17}+1}{2}}$
$x_{4}=-(\sqrt{\dfrac{\sqrt{17}-1}{2}}-i\sqrt{\dfrac{\sqrt{17}+1}{2}})$

$\text{Ответ:}$

$$
\begin{array}{l}
x_{1}=\sqrt{\dfrac{\sqrt{17}-1}{2}}+i\sqrt{\dfrac{\sqrt{17}+1}{2}}\\
x_{2}=-(\sqrt{\dfrac{\sqrt{17}-1}{2}}+i\sqrt{\dfrac{\sqrt{17}+1}{2}})\\
x_{3}=\sqrt{\dfrac{\sqrt{17}-1}{2}}-i\sqrt{\dfrac{\sqrt{17}+1}{2}}\\
x_{4}=-(\sqrt{\dfrac{\sqrt{17}-1}{2}}-i\sqrt{\dfrac{\sqrt{17}+1}{2}})
\end{array}
$$

## 35 номер

### Пример:

$x^4 + 10x^2 + 61 = 0$

### Решение:

$t=x^{2}$
$t^2+10t+61=0$
$D-144$
$t=\dfrac{-10\pm 12i}{2}=-5\pm 6i$

$x^2=-5\pm 6i$

$\text{1. }x^2=-5+6i$
$x=a+bi$
$(a+bi)^2=(a^2-b^2)+2abi=-5+6i$

$$
\begin{cases}
a^2-b^2=-5 \\
2ab=6
\end{cases}
$$

$(a^2+b^2)^2=(a^2-b^2)^2+(2ab)^2=(-5)^2+6^2=61$
$a^2+b^2=\sqrt{61}$
$a^2=\dfrac{(a^2+b^2)+(a^2-b^2)}{2}=\dfrac{\sqrt{61}-5}{2}$
$b^2=\dfrac{(a^2+b^2)-(a^2-b^2)}{2}=\dfrac{\sqrt{61}+5}{2}$

$x_{1}=\sqrt{\dfrac{\sqrt{61}-5}{2}}+i\sqrt{\dfrac{\sqrt{61}+5}{2}}$
$x_{2}=-(\sqrt{\dfrac{\sqrt{61}-5}{2}}+i\sqrt{\dfrac{\sqrt{61}+5}{2}})$

$\text{2. }x^2=-5-6i$
$-5-6i=\overline{(-5+6i)}$

$x_{3}=\sqrt{\dfrac{\sqrt{61}-5}{2}}-i\sqrt{\dfrac{\sqrt{61}+5}{2}}$
$x_{4}=-(\sqrt{\dfrac{\sqrt{61}-5}{2}}-i\sqrt{\dfrac{\sqrt{61}+5}{2}})$

$\text{Ответ:}$

$$
\begin{array}{l}
x_{1}=\sqrt{\dfrac{\sqrt{61}-5}{2}}+i\sqrt{\dfrac{\sqrt{61}+5}{2}}\\
x_{2}=-(\sqrt{\dfrac{\sqrt{61}-5}{2}}+i\sqrt{\dfrac{\sqrt{61}+5}{2}})\\
x_{3}=\sqrt{\dfrac{\sqrt{61}-5}{2}}-i\sqrt{\dfrac{\sqrt{61}+5}{2}}\\
x_{4}=-(\sqrt{\dfrac{\sqrt{61}-5}{2}}-i\sqrt{\dfrac{\sqrt{61}+5}{2}})
\end{array}
$$

## 36 номер

### Пример:

$x^4 − x^2 + 37 = 0$

### Решение:

$t=x^{2}$
$t^2-t+37=0$
$D-147$
$t=\dfrac{1\pm\sqrt{-147}}{2}=\dfrac{1\pm 7i\sqrt3}{2}$

$x^2=\dfrac{1\pm 7i\sqrt3}{2}$

$\text{1. }x^2=\dfrac{1+7i\sqrt3}{2}$
$x=a+bi$
$(a+bi)^2=(a^2-b^2)+2abi=\dfrac{1+7i\sqrt3}{2}$

$$
\begin{cases}
a^2-b^2=\dfrac{1}{2} \\
2ab=\dfrac{7\sqrt3}{2}
\end{cases}
$$

$(a^2+b^2)^2=(a^2-b^2)^2+(2ab)^2=(\dfrac{1}{2})^2+(\dfrac{7\sqrt3}{2})^2=\dfrac{1}{4}+\dfrac{147}{4}=37$
$a^2+b^2=\sqrt{37}$

$a^2=\dfrac{(a^2+b^2)+(a^2-b^2)}{2}=\dfrac{\sqrt{37}+\frac{1}{2}}{2}=\dfrac{2\sqrt{37}+1}{4}$

$b^2=\dfrac{(a^2+b^2)-(a^2-b^2)}{2}=\dfrac{\sqrt{37}-\frac{1}{2}}{2}=\dfrac{2\sqrt{37}-1}{4}$

$x_{1}=\sqrt{\dfrac{2\sqrt{37}+1}{4}}+i\sqrt{\dfrac{2\sqrt{37}-1}{4}}$
$x_{2}=-(\sqrt{\dfrac{2\sqrt{37}+1}{4}}+i\sqrt{\dfrac{2\sqrt{37}-1}{4}})$

$\text{2. }x^2=\dfrac{1-7i\sqrt3}{2}$
$\dfrac{1-7i\sqrt3}{2}=\overline{(\dfrac{1+7i\sqrt3}{2})}$

$x_{3}=\sqrt{\dfrac{2\sqrt{37}+1}{4}}-i\sqrt{\dfrac{2\sqrt{37}-1}{4}}$
$x_{4}=-(\sqrt{\dfrac{2\sqrt{37}+1}{4}}-i\sqrt{\dfrac{2\sqrt{37}-1}{4}})$

$\text{Ответ:}$

$$
\begin{array}{l}
x_{1}=\sqrt{\dfrac{2\sqrt{37}+1}{4}}+i\sqrt{\dfrac{2\sqrt{37}-1}{4}}\\
x_{2}=-(\sqrt{\dfrac{2\sqrt{37}+1}{4}}+i\sqrt{\dfrac{2\sqrt{37}-1}{4}})\\
x_{3}=\sqrt{\dfrac{2\sqrt{37}+1}{4}}-i\sqrt{\dfrac{2\sqrt{37}-1}{4}}\\
x_{4}=-(\sqrt{\dfrac{2\sqrt{37}+1}{4}}-i\sqrt{\dfrac{2\sqrt{37}-1}{4}})
\end{array}
$$

## 37 номер

### Пример:

$x^4 + 6x^2 + 8 = 0$

### Решение:

$t=x^{2}$
$t^2+6t+8=0$
$D=6^2-4\cdot1\cdot8=36-32=4$
$t=\dfrac{-6\pm\sqrt{4}}{2}=-3\pm1$

$t_1=-2,\quad t_2=-4$

$x^2=-2\ \ \text{or}\ \ x^2=-4$

$\text{1. }x^2=-2$
$x=\pm\sqrt{-2}=\pm i\sqrt2$
$x_{1}=i\sqrt2$
$x_{2}=-i\sqrt2$

$\text{2. }x^2=-4$
$x=\pm\sqrt{-4}=\pm 2i$
$x_{3}=2i$
$x_{4}=-2i$

$\text{Ответ:}$

$$
\begin{array}{l}
x_{1}=i\sqrt2\\
x_{2}=-i\sqrt2\\
x_{3}=2i\\
x_{4}=-2i
\end{array}
$$

## 38 номер

### Пример:

$x4 + 8x^2 + 41 = 0$

### Решение:

$t=x^{2}$
$t^2+8t+41=0$
$D=8^2-4\cdot1\cdot41=64-164=-100$
$t=\dfrac{-8\pm\sqrt{-100}}{2}=\dfrac{-8\pm 10i}{2}=-4\pm 5i$

$x^2=-4\pm 5i$

$\text{1. }x^2=-4+5i$
$x=a+bi$
$(a+bi)^2=(a^2-b^2)+2abi=-4+5i$

$$
\begin{cases}
a^2-b^2=-4 \\
2ab=5
\end{cases}
$$

$(a^2+b^2)^2=(a^2-b^2)^2+(2ab)^2=(-4)^2+5^2=41$
$a^2+b^2=\sqrt{41}$
$a^2=\dfrac{(a^2+b^2)+(a^2-b^2)}{2}=\dfrac{\sqrt{41}-4}{2}$
$b^2=\dfrac{(a^2+b^2)-(a^2-b^2)}{2}=\dfrac{\sqrt{41}+4}{2}$

$x_{1}=\sqrt{\dfrac{\sqrt{41}-4}{2}}+i\sqrt{\dfrac{\sqrt{41}+4}{2}}$
$x_{2}=-(\sqrt{\dfrac{\sqrt{41}-4}{2}}+i\sqrt{\dfrac{\sqrt{41}+4}{2}})$

$\text{2. }x^2=-4-5i$
$-4-5i=\overline{(-4+5i)}$

$x_{3}=\sqrt{\dfrac{\sqrt{41}-4}{2}}-i\sqrt{\dfrac{\sqrt{41}+4}{2}}$
$x_{4}=-(\sqrt{\dfrac{\sqrt{41}-4}{2}}-i\sqrt{\dfrac{\sqrt{41}+4}{2}})$

$\text{Ответ:}$

$$
\begin{array}{l}
x_{1}=\sqrt{\dfrac{\sqrt{41}-4}{2}}+i\sqrt{\dfrac{\sqrt{41}+4}{2}}\\
x_{2}=-(\sqrt{\dfrac{\sqrt{41}-4}{2}}+i\sqrt{\dfrac{\sqrt{41}+4}{2}})\\
x_{3}=\sqrt{\dfrac{\sqrt{41}-4}{2}}-i\sqrt{\dfrac{\sqrt{41}+4}{2}}\\
x_{4}=-(\sqrt{\dfrac{\sqrt{41}-4}{2}}-i\sqrt{\dfrac{\sqrt{41}+4}{2}})
\end{array}
$$

## 39 номер

### Условие:

$z-i\leq 1$

### Область:

![[Pasted image 20251225172031.png]]

## 40 номер

### Условие:

$\mathrm{Re}(z)\leq 3$

### Область:

![[Pasted image 20251225172459.png]]

## 41 номер

### Условие:

$z\leq 2 \ \ \text{and} \ \ \mathrm{Re}(z)\geq 0$

### Область:

![[Pasted image 20251225173000.png]]

## 42 номер

### Условие:

$arg(z)\leq \dfrac{\pi}{6}$

### Область:

![[Pasted image 20251225174710.png]]

## 43 номер

### Условие:
$z=5;arg(z)\leq \dfrac{\pi}{3}$

### Область:
![[IMG_0055.jpeg]]

## 44 номер

### Пример:
$z+i\leq1; \mathrm{Im}\leq -1$

### Решение:
![[IMG_0056.jpeg]]