468 lines
9.4 KiB
Markdown
468 lines
9.4 KiB
Markdown
## 30 номер
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#семестр_1 #высшая_математика
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### Пример:
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$x^3+1=0$
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### Решение:
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$x^3+1=(x+1)(x^2-x+1)$
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$x = -1;$
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$x^2-x+1=0$
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$D = -3$
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$\text{Ответ: } x=\dfrac{1\pm i\sqrt{ 3 }}{2}$
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## 31 номер
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### Пример:
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$x^4-4x^2+5=0$
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### Решение:
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$t=x^{2}$
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$t^2-4t+5=0$
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$D=-4$
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$t=\dfrac{4\pm 2i}{2}=2\pm i$
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$x^2=2\pm i$
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$\text{1. }x^2=2+i$
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$x=a+bi$
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$(a+bi)^2=(a^2-b^2)+2abi=2+i$
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$$
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\begin{cases}
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a^2-b^2=2 \\
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2abi=i; 2ab=1
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\end{cases}
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$$
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$(a^2+b^2)^2=(a^2-b^2)^2+(2ab)^2=2^2+1^2=5$
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$a^2+b^2=\sqrt{5}$
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$a^2=\dfrac{(a^2+b^2)+(a^2-b^2)}{2}=\dfrac{\sqrt{ 5 }+2}{2}$
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$b^2=\dfrac{(a^2+b^2)-(a^2-b^2)}{2}=\dfrac{\sqrt{ 5 }-2}{2}$
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$x_{1}=\sqrt{ \dfrac{\sqrt{ 5 }+2}{2} }+i\sqrt{ \dfrac{\sqrt{ 5 }-2}{2} }$
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$x_{2}=-( \sqrt{ \dfrac{\sqrt{ 5 }+2}{2} }+i\sqrt{ \dfrac{\sqrt{ 5 }-2}{2} } )$
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$\text{2. }x^2=2-i$
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$2-i=\overline{(2+i)}$
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$x_{3}=\sqrt{ \dfrac{\sqrt{ 5 }+2}{2} }-i\sqrt{ \dfrac{\sqrt{ 5 }-2}{2} }$
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$x_{4}=-( \sqrt{ \dfrac{\sqrt{ 5 }+2}{2} }-i\sqrt{ \dfrac{\sqrt{ 5 }-2}{2} } )$
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$\text{Ответ:}$
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$$
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\begin{array} \\
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x_{1}=\sqrt{ \dfrac{\sqrt{ 5 }+2}{2} }+i\sqrt{ \dfrac{\sqrt{ 5 }-2}{2} } \\
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x_{2}=-( \sqrt{ \dfrac{\sqrt{ 5 }+2}{2} }+i\sqrt{ \dfrac{\sqrt{ 5 }-2}{2} } ) \\
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x_{3}=\sqrt{ \dfrac{\sqrt{ 5 }+2}{2} }-i\sqrt{ \dfrac{\sqrt{ 5 }-2}{2} } \\
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x_{4}=-( \sqrt{ \dfrac{\sqrt{ 5 }+2}{2} }-i\sqrt{ \dfrac{\sqrt{ 5 }-2}{2} } )
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\end{array}
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$$
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## 32 номер
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### Пример:
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$x^4+4x^2+20=0$
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### Решение:
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$t=x^{2}$
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$t^2+4t+20=0$
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$D=-64$
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$t=\dfrac{-4\pm 8i}{2}=-2\pm 4i$
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$x^2=-2\pm 4i$
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$\text{1. }x^2=-2+4i$
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$x=a+bi$
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$(a+bi)^2=(a^2-b^2)+2abi=-2+4i$
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$$
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\begin{cases}
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a^2-b^2=-2 \\
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2ab=4
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\end{cases}
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$$
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$(a^2+b^2)^2=(a^2-b^2)^2+(2ab)^2=(-2)^2+4^2=20$
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$a^2+b^2=\sqrt{20}=2\sqrt5$
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$a^2=\dfrac{(a^2+b^2)+(a^2-b^2)}{2}=\dfrac{2\sqrt5-2}{2}=\sqrt5-1$
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$b^2=\dfrac{(a^2+b^2)-(a^2-b^2)}{2}=\dfrac{2\sqrt5+2}{2}=\sqrt5+1$
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$x_{1}=\sqrt{\sqrt5-1}+i\sqrt{\sqrt5+1}$
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$x_{2}=-(\sqrt{\sqrt5-1}+i\sqrt{\sqrt5+1})$
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$\text{2. }x^2=-2-4i$
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$-2-4i=\overline{(-2+4i)}$
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$x_{3}=\sqrt{\sqrt5-1}-i\sqrt{\sqrt5+1}$
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$x_{4}=-(\sqrt{\sqrt5-1}-i\sqrt{\sqrt5+1})$
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$\text{Ответ:}$
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$$
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\begin{array}{l}
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x_{1}=\sqrt{\sqrt5-1}+i\sqrt{\sqrt5+1}\\
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x_{2}=-(\sqrt{\sqrt5-1}+i\sqrt{\sqrt5+1})\\
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x_{3}=\sqrt{\sqrt5-1}-i\sqrt{\sqrt5+1}\\
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x_{4}=-(\sqrt{\sqrt5-1}-i\sqrt{\sqrt5+1})
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\end{array}
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$$
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## 33 номер
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### Пример:
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$x^4-6x^2+13=0$
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### Решение:
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$t=x^{2}$
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$t^2-6t+13=0$
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$D=-16$
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$t=\dfrac{6\pm4i}{2}=3\pm2i$
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$x^2=3\pm2i$
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$\text{1. }x^2=3+2i$
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$x=a+bi$
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$(a+bi)^2=(a^2-b^2)+2abi=3+2i$
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$$
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\begin{cases}
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a^2-b^2=3 \\
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2ab=2
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\end{cases}
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$$
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$(a^2+b^2)^2=(a^2-b^2)^2+(2ab)^2=3^2+2^2=13$
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$a^2+b^2=\sqrt{13}$
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$a^2=\dfrac{(a^2+b^2)+(a^2-b^2)}{2}=\dfrac{\sqrt{13}+3}{2}$
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$b^2=\dfrac{(a^2+b^2)-(a^2-b^2)}{2}=\dfrac{\sqrt{13}-3}{2}$
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$x_{1}=\sqrt{\dfrac{\sqrt{13}+3}{2}}+i\sqrt{\dfrac{\sqrt{13}-3}{2}}$
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$x_{2}=-(\sqrt{\dfrac{\sqrt{13}+3}{2}}+i\sqrt{\dfrac{\sqrt{13}-3}{2}})$
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$\text{2. }x^2=3-2i$
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$3-2i=\overline{(3+2i)}$
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$x_{3}=\sqrt{\dfrac{\sqrt{13}+3}{2}}-i\sqrt{\dfrac{\sqrt{13}-3}{2}}$
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$x_{4}=-(\sqrt{\dfrac{\sqrt{13}+3}{2}}-i\sqrt{\dfrac{\sqrt{13}-3}{2}})$
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$\text{Ответ:}$
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$$
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\begin{array}{l}
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x_{1}=\sqrt{\dfrac{\sqrt{13}+3}{2}}+i\sqrt{\dfrac{\sqrt{13}-3}{2}}\\
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x_{2}=-(\sqrt{\dfrac{\sqrt{13}+3}{2}}+i\sqrt{\dfrac{\sqrt{13}-3}{2}})\\
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x_{3}=\sqrt{\dfrac{\sqrt{13}+3}{2}}-i\sqrt{\dfrac{\sqrt{13}-3}{2}}\\
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x_{4}=-(\sqrt{\dfrac{\sqrt{13}+3}{2}}-i\sqrt{\dfrac{\sqrt{13}-3}{2}})
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\end{array}
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$$
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## 34 номер
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### Пример:
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$x^4 + 2x^2 + 17 = 0$
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### Решение:
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$t=x^{2}$
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$t^2+2t+17=0$
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$D=-64$
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$t=\dfrac{-2\pm 8i}{2}=-1\pm 4i$
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$x^2=-1\pm 4i$
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$\text{1. }x^2=-1+4i$
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$x=a+bi$
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$(a+bi)^2=(a^2-b^2)+2abi=-1+4i$
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$$
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\begin{cases}
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a^2-b^2=-1 \\
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2ab=4
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\end{cases}
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$$
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$(a^2+b^2)^2=(a^2-b^2)^2+(2ab)^2=(-1)^2+4^2=17$
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$a^2+b^2=\sqrt{17}$
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$a^2=\dfrac{(a^2+b^2)+(a^2-b^2)}{2}=\dfrac{\sqrt{17}-1}{2}$
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$b^2=\dfrac{(a^2+b^2)-(a^2-b^2)}{2}=\dfrac{\sqrt{17}+1}{2}$
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$x_{1}=\sqrt{\dfrac{\sqrt{17}-1}{2}}+i\sqrt{\dfrac{\sqrt{17}+1}{2}}$
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$x_{2}=-(\sqrt{\dfrac{\sqrt{17}-1}{2}}+i\sqrt{\dfrac{\sqrt{17}+1}{2}})$
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$\text{2. }x^2=-1-4i$
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$-1-4i=\overline{(-1+4i)}$
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$x_{3}=\sqrt{\dfrac{\sqrt{17}-1}{2}}-i\sqrt{\dfrac{\sqrt{17}+1}{2}}$
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$x_{4}=-(\sqrt{\dfrac{\sqrt{17}-1}{2}}-i\sqrt{\dfrac{\sqrt{17}+1}{2}})$
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$\text{Ответ:}$
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$$
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\begin{array}{l}
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x_{1}=\sqrt{\dfrac{\sqrt{17}-1}{2}}+i\sqrt{\dfrac{\sqrt{17}+1}{2}}\\
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x_{2}=-(\sqrt{\dfrac{\sqrt{17}-1}{2}}+i\sqrt{\dfrac{\sqrt{17}+1}{2}})\\
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x_{3}=\sqrt{\dfrac{\sqrt{17}-1}{2}}-i\sqrt{\dfrac{\sqrt{17}+1}{2}}\\
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x_{4}=-(\sqrt{\dfrac{\sqrt{17}-1}{2}}-i\sqrt{\dfrac{\sqrt{17}+1}{2}})
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\end{array}
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$$
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## 35 номер
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### Пример:
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$x^4 + 10x^2 + 61 = 0$
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### Решение:
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$t=x^{2}$
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$t^2+10t+61=0$
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$D-144$
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$t=\dfrac{-10\pm 12i}{2}=-5\pm 6i$
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$x^2=-5\pm 6i$
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$\text{1. }x^2=-5+6i$
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$x=a+bi$
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$(a+bi)^2=(a^2-b^2)+2abi=-5+6i$
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$$
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\begin{cases}
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a^2-b^2=-5 \\
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2ab=6
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\end{cases}
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$$
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$(a^2+b^2)^2=(a^2-b^2)^2+(2ab)^2=(-5)^2+6^2=61$
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$a^2+b^2=\sqrt{61}$
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$a^2=\dfrac{(a^2+b^2)+(a^2-b^2)}{2}=\dfrac{\sqrt{61}-5}{2}$
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$b^2=\dfrac{(a^2+b^2)-(a^2-b^2)}{2}=\dfrac{\sqrt{61}+5}{2}$
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$x_{1}=\sqrt{\dfrac{\sqrt{61}-5}{2}}+i\sqrt{\dfrac{\sqrt{61}+5}{2}}$
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$x_{2}=-(\sqrt{\dfrac{\sqrt{61}-5}{2}}+i\sqrt{\dfrac{\sqrt{61}+5}{2}})$
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$\text{2. }x^2=-5-6i$
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$-5-6i=\overline{(-5+6i)}$
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$x_{3}=\sqrt{\dfrac{\sqrt{61}-5}{2}}-i\sqrt{\dfrac{\sqrt{61}+5}{2}}$
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$x_{4}=-(\sqrt{\dfrac{\sqrt{61}-5}{2}}-i\sqrt{\dfrac{\sqrt{61}+5}{2}})$
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$\text{Ответ:}$
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$$
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\begin{array}{l}
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x_{1}=\sqrt{\dfrac{\sqrt{61}-5}{2}}+i\sqrt{\dfrac{\sqrt{61}+5}{2}}\\
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x_{2}=-(\sqrt{\dfrac{\sqrt{61}-5}{2}}+i\sqrt{\dfrac{\sqrt{61}+5}{2}})\\
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x_{3}=\sqrt{\dfrac{\sqrt{61}-5}{2}}-i\sqrt{\dfrac{\sqrt{61}+5}{2}}\\
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x_{4}=-(\sqrt{\dfrac{\sqrt{61}-5}{2}}-i\sqrt{\dfrac{\sqrt{61}+5}{2}})
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\end{array}
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$$
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## 36 номер
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### Пример:
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$x^4 − x^2 + 37 = 0$
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### Решение:
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$t=x^{2}$
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$t^2-t+37=0$
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$D-147$
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$t=\dfrac{1\pm\sqrt{-147}}{2}=\dfrac{1\pm 7i\sqrt3}{2}$
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$x^2=\dfrac{1\pm 7i\sqrt3}{2}$
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$\text{1. }x^2=\dfrac{1+7i\sqrt3}{2}$
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$x=a+bi$
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$(a+bi)^2=(a^2-b^2)+2abi=\dfrac{1+7i\sqrt3}{2}$
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$$
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\begin{cases}
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a^2-b^2=\dfrac{1}{2} \\
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2ab=\dfrac{7\sqrt3}{2}
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\end{cases}
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$$
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$(a^2+b^2)^2=(a^2-b^2)^2+(2ab)^2=(\dfrac{1}{2})^2+(\dfrac{7\sqrt3}{2})^2=\dfrac{1}{4}+\dfrac{147}{4}=37$
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$a^2+b^2=\sqrt{37}$
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$a^2=\dfrac{(a^2+b^2)+(a^2-b^2)}{2}=\dfrac{\sqrt{37}+\frac{1}{2}}{2}=\dfrac{2\sqrt{37}+1}{4}$
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$b^2=\dfrac{(a^2+b^2)-(a^2-b^2)}{2}=\dfrac{\sqrt{37}-\frac{1}{2}}{2}=\dfrac{2\sqrt{37}-1}{4}$
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$x_{1}=\sqrt{\dfrac{2\sqrt{37}+1}{4}}+i\sqrt{\dfrac{2\sqrt{37}-1}{4}}$
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$x_{2}=-(\sqrt{\dfrac{2\sqrt{37}+1}{4}}+i\sqrt{\dfrac{2\sqrt{37}-1}{4}})$
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$\text{2. }x^2=\dfrac{1-7i\sqrt3}{2}$
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$\dfrac{1-7i\sqrt3}{2}=\overline{(\dfrac{1+7i\sqrt3}{2})}$
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$x_{3}=\sqrt{\dfrac{2\sqrt{37}+1}{4}}-i\sqrt{\dfrac{2\sqrt{37}-1}{4}}$
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$x_{4}=-(\sqrt{\dfrac{2\sqrt{37}+1}{4}}-i\sqrt{\dfrac{2\sqrt{37}-1}{4}})$
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$\text{Ответ:}$
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$$
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\begin{array}{l}
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x_{1}=\sqrt{\dfrac{2\sqrt{37}+1}{4}}+i\sqrt{\dfrac{2\sqrt{37}-1}{4}}\\
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x_{2}=-(\sqrt{\dfrac{2\sqrt{37}+1}{4}}+i\sqrt{\dfrac{2\sqrt{37}-1}{4}})\\
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x_{3}=\sqrt{\dfrac{2\sqrt{37}+1}{4}}-i\sqrt{\dfrac{2\sqrt{37}-1}{4}}\\
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x_{4}=-(\sqrt{\dfrac{2\sqrt{37}+1}{4}}-i\sqrt{\dfrac{2\sqrt{37}-1}{4}})
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\end{array}
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$$
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## 37 номер
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### Пример:
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$x^4 + 6x^2 + 8 = 0$
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### Решение:
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$t=x^{2}$
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$t^2+6t+8=0$
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$D=6^2-4\cdot1\cdot8=36-32=4$
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$t=\dfrac{-6\pm\sqrt{4}}{2}=-3\pm1$
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$t_1=-2,\quad t_2=-4$
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$x^2=-2\ \ \text{or}\ \ x^2=-4$
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$\text{1. }x^2=-2$
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$x=\pm\sqrt{-2}=\pm i\sqrt2$
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$x_{1}=i\sqrt2$
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$x_{2}=-i\sqrt2$
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$\text{2. }x^2=-4$
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$x=\pm\sqrt{-4}=\pm 2i$
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$x_{3}=2i$
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$x_{4}=-2i$
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$\text{Ответ:}$
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$$
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\begin{array}{l}
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x_{1}=i\sqrt2\\
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x_{2}=-i\sqrt2\\
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x_{3}=2i\\
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x_{4}=-2i
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\end{array}
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$$
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## 38 номер
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### Пример:
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$x4 + 8x^2 + 41 = 0$
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### Решение:
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$t=x^{2}$
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$t^2+8t+41=0$
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$D=8^2-4\cdot1\cdot41=64-164=-100$
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$t=\dfrac{-8\pm\sqrt{-100}}{2}=\dfrac{-8\pm 10i}{2}=-4\pm 5i$
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$x^2=-4\pm 5i$
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$\text{1. }x^2=-4+5i$
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$x=a+bi$
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$(a+bi)^2=(a^2-b^2)+2abi=-4+5i$
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$$
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\begin{cases}
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a^2-b^2=-4 \\
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2ab=5
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\end{cases}
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$$
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$(a^2+b^2)^2=(a^2-b^2)^2+(2ab)^2=(-4)^2+5^2=41$
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$a^2+b^2=\sqrt{41}$
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$a^2=\dfrac{(a^2+b^2)+(a^2-b^2)}{2}=\dfrac{\sqrt{41}-4}{2}$
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$b^2=\dfrac{(a^2+b^2)-(a^2-b^2)}{2}=\dfrac{\sqrt{41}+4}{2}$
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$x_{1}=\sqrt{\dfrac{\sqrt{41}-4}{2}}+i\sqrt{\dfrac{\sqrt{41}+4}{2}}$
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$x_{2}=-(\sqrt{\dfrac{\sqrt{41}-4}{2}}+i\sqrt{\dfrac{\sqrt{41}+4}{2}})$
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$\text{2. }x^2=-4-5i$
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$-4-5i=\overline{(-4+5i)}$
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$x_{3}=\sqrt{\dfrac{\sqrt{41}-4}{2}}-i\sqrt{\dfrac{\sqrt{41}+4}{2}}$
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$x_{4}=-(\sqrt{\dfrac{\sqrt{41}-4}{2}}-i\sqrt{\dfrac{\sqrt{41}+4}{2}})$
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$\text{Ответ:}$
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$$
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\begin{array}{l}
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x_{1}=\sqrt{\dfrac{\sqrt{41}-4}{2}}+i\sqrt{\dfrac{\sqrt{41}+4}{2}}\\
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x_{2}=-(\sqrt{\dfrac{\sqrt{41}-4}{2}}+i\sqrt{\dfrac{\sqrt{41}+4}{2}})\\
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x_{3}=\sqrt{\dfrac{\sqrt{41}-4}{2}}-i\sqrt{\dfrac{\sqrt{41}+4}{2}}\\
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x_{4}=-(\sqrt{\dfrac{\sqrt{41}-4}{2}}-i\sqrt{\dfrac{\sqrt{41}+4}{2}})
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\end{array}
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$$
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## 39 номер
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### Условие:
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$z-i\leq 1$
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### Область:
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![[Pasted image 20251225172031.png]]
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## 40 номер
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### Условие:
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$\mathrm{Re}(z)\leq 3$
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### Область:
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![[Pasted image 20251225172459.png]]
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## 41 номер
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### Условие:
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$z\leq 2 \ \ \text{and} \ \ \mathrm{Re}(z)\geq 0$
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### Область:
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![[Pasted image 20251225173000.png]]
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## 42 номер
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### Условие:
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$arg(z)\leq \dfrac{\pi}{6}$
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### Область:
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![[Pasted image 20251225174710.png]]
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## 43 номер
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### Условие:
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$z=5;arg(z)\leq \dfrac{\pi}{3}$
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### Область:
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![[IMG_0055.jpeg]]
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## 44 номер
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### Пример:
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$z+i\leq1; \mathrm{Im}\leq -1$
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### Решение:
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![[IMG_0056.jpeg]]
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