x^4-y=-2 (уравнение)

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    Найду корень уравнения: x^4-y=-2

    Решение

    Вы ввели [src]
     4         
    x  - y = -2
    x4y=2x^{4} - y = -2
    Быстрый ответ [src]
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         8 /             2     2        /atan2(im(y), -2 + re(y))\     8 /             2     2        /atan2(im(y), -2 + re(y))\
    x1 = \/  (-2 + re(y))  + im (y) *sin|------------------------| - I*\/  (-2 + re(y))  + im (y) *cos|------------------------|
                                        \           4            /                                    \           4            /
    x1=(y2)2+(y)28sin(14atan2(y,y2))i(y2)2+(y)28cos(14atan2(y,y2))x_{1} = \sqrt[8]{\left(\Re{y} - 2\right)^{2} + \left(\Im{y}\right)^{2}} \sin{\left (\frac{1}{4} \operatorname{atan_{2}}{\left (\Im{y},\Re{y} - 2 \right )} \right )} - i \sqrt[8]{\left(\Re{y} - 2\right)^{2} + \left(\Im{y}\right)^{2}} \cos{\left (\frac{1}{4} \operatorname{atan_{2}}{\left (\Im{y},\Re{y} - 2 \right )} \right )}
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           8 /             2     2        /atan2(im(y), -2 + re(y))\     8 /             2     2        /atan2(im(y), -2 + re(y))\
    x2 = - \/  (-2 + re(y))  + im (y) *sin|------------------------| + I*\/  (-2 + re(y))  + im (y) *cos|------------------------|
                                          \           4            /                                    \           4            /
    x2=(y2)2+(y)28sin(14atan2(y,y2))+i(y2)2+(y)28cos(14atan2(y,y2))x_{2} = - \sqrt[8]{\left(\Re{y} - 2\right)^{2} + \left(\Im{y}\right)^{2}} \sin{\left (\frac{1}{4} \operatorname{atan_{2}}{\left (\Im{y},\Re{y} - 2 \right )} \right )} + i \sqrt[8]{\left(\Re{y} - 2\right)^{2} + \left(\Im{y}\right)^{2}} \cos{\left (\frac{1}{4} \operatorname{atan_{2}}{\left (\Im{y},\Re{y} - 2 \right )} \right )}
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           8 /             2     2        /atan2(im(y), -2 + re(y))\     8 /             2     2        /atan2(im(y), -2 + re(y))\
    x3 = - \/  (-2 + re(y))  + im (y) *cos|------------------------| - I*\/  (-2 + re(y))  + im (y) *sin|------------------------|
                                          \           4            /                                    \           4            /
    x3=i(y2)2+(y)28sin(14atan2(y,y2))(y2)2+(y)28cos(14atan2(y,y2))x_{3} = - i \sqrt[8]{\left(\Re{y} - 2\right)^{2} + \left(\Im{y}\right)^{2}} \sin{\left (\frac{1}{4} \operatorname{atan_{2}}{\left (\Im{y},\Re{y} - 2 \right )} \right )} - \sqrt[8]{\left(\Re{y} - 2\right)^{2} + \left(\Im{y}\right)^{2}} \cos{\left (\frac{1}{4} \operatorname{atan_{2}}{\left (\Im{y},\Re{y} - 2 \right )} \right )}
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         8 /             2     2        /atan2(im(y), -2 + re(y))\     8 /             2     2        /atan2(im(y), -2 + re(y))\
    x4 = \/  (-2 + re(y))  + im (y) *cos|------------------------| + I*\/  (-2 + re(y))  + im (y) *sin|------------------------|
                                        \           4            /                                    \           4            /
    x4=i(y2)2+(y)28sin(14atan2(y,y2))+(y2)2+(y)28cos(14atan2(y,y2))x_{4} = i \sqrt[8]{\left(\Re{y} - 2\right)^{2} + \left(\Im{y}\right)^{2}} \sin{\left (\frac{1}{4} \operatorname{atan_{2}}{\left (\Im{y},\Re{y} - 2 \right )} \right )} + \sqrt[8]{\left(\Re{y} - 2\right)^{2} + \left(\Im{y}\right)^{2}} \cos{\left (\frac{1}{4} \operatorname{atan_{2}}{\left (\Im{y},\Re{y} - 2 \right )} \right )}