Опустим высоту PH
треугольник MPH прямоугольный.
cosM=3/4
cosM=mh/mp
mh=9
т.к MPH- равнобедренный PH-бис. и медиана след MK= 18.
Ответ: MK=18
Выражаем из первого
=>
![\left \{ {{2 \sqrt[3]{x} = 4 } \atop {{ \sqrt[3]{y}= 1 \right. =\ \textgreater \ \left \{ {{ \sqrt[3]{x} = 2 } \atop {{ \sqrt[3]{y}= 1 \right. =\ \textgreater \ \left \{ {{{x} = 2^{3} } \atop {{ {y}= 1 \right.](https://tex.z-dn.net/?f=%5Cleft+%5C%7B+%7B%7B2+%5Csqrt%5B3%5D%7Bx%7D+%3D+4+%7D+%5Catop+%7B%7B+%5Csqrt%5B3%5D%7By%7D%3D+1+%5Cright.+%3D%5C+%5Ctextgreater+%5C++%5Cleft+%5C%7B+%7B%7B+%5Csqrt%5B3%5D%7Bx%7D+%3D+2+%7D+%5Catop+%7B%7B+%5Csqrt%5B3%5D%7By%7D%3D+1+%5Cright.+%3D%5C+%5Ctextgreater+%5C++%5Cleft+%5C%7B+%7B%7B%7Bx%7D+%3D+2%5E%7B3%7D+%7D+%5Catop+%7B%7B+%7By%7D%3D+1+%5Cright.)
Второй листок.
![\left \{ {{x^2 - 3 \sqrt[5]{y}=5 } \atop {\sqrt[5]{y} = x+1}} \right. =\ \textgreater \ \left \{ {{x^2 - 3(x+1)=5 } \atop {\sqrt[5]{y} = x+1}} \right. =\ \textgreater \ \left \{ {{x^2 - 3x-2=0 } \atop {\sqrt[5]{y} = x+1}} \right.](https://tex.z-dn.net/?f=+%5Cleft+%5C%7B+%7B%7Bx%5E2+-+3+%5Csqrt%5B5%5D%7By%7D%3D5+%7D+%5Catop+%7B%5Csqrt%5B5%5D%7By%7D+%3D+x%2B1%7D%7D+%5Cright.+%3D%5C+%5Ctextgreater+%5C++++%5Cleft+%5C%7B+%7B%7Bx%5E2+-+3%28x%2B1%29%3D5+%7D+%5Catop+%7B%5Csqrt%5B5%5D%7By%7D+%3D+x%2B1%7D%7D+%5Cright.+%3D%5C+%5Ctextgreater+%5C+++%5Cleft+%5C%7B+%7B%7Bx%5E2+-+3x-2%3D0+%7D+%5Catop+%7B%5Csqrt%5B5%5D%7By%7D+%3D+x%2B1%7D%7D+%5Cright.+)
Решаем квадратное уравнение в 1 уравнении. Получаем корни х = 2 и х = 1.
Рассматриваем два случая:
![\left \{ {{x=2} \atop {\sqrt[5]{y}=x+1}} \right. =\ \textgreater \ \left \{ {{x=2} \atop {\sqrt[5]{y}=3}} \right. =\ \textgreater \ \left \{ {{x=2} \atop {y= 5^3=125}} \right.](https://tex.z-dn.net/?f=+%5Cleft+%5C%7B+%7B%7Bx%3D2%7D+%5Catop+%7B%5Csqrt%5B5%5D%7By%7D%3Dx%2B1%7D%7D+%5Cright.+%3D%5C+%5Ctextgreater+%5C+++%5Cleft+%5C%7B+%7B%7Bx%3D2%7D+%5Catop+%7B%5Csqrt%5B5%5D%7By%7D%3D3%7D%7D+%5Cright.++%3D%5C+%5Ctextgreater+%5C++%5Cleft+%5C%7B+%7B%7Bx%3D2%7D+%5Catop+%7By%3D+5%5E3%3D125%7D%7D+%5Cright.+)
![\left \{ {{x=1} \atop {\sqrt[5]{y}=x+1}} \right. =\ \textgreater \ \left \{ {{x=1} \atop {\sqrt[5]{y}=2}} \right. =\ \textgreater \ \left \{ {{x=1} \atop {y=2^5=32}} \right.](https://tex.z-dn.net/?f=+%5Cleft+%5C%7B+%7B%7Bx%3D1%7D+%5Catop+%7B%5Csqrt%5B5%5D%7By%7D%3Dx%2B1%7D%7D+%5Cright.+%3D%5C+%5Ctextgreater+%5C++%5Cleft+%5C%7B+%7B%7Bx%3D1%7D+%5Catop+%7B%5Csqrt%5B5%5D%7By%7D%3D2%7D%7D+%5Cright.+%3D%5C+%5Ctextgreater+%5C++%5Cleft+%5C%7B+%7B%7Bx%3D1%7D+%5Catop+%7By%3D2%5E5%3D32%7D%7D+%5Cright.++)
Рассмотрим комплексные числа
и
. Тогда их модуль комплексных чисел:
![|a|=\sqrt{(\sqrt{2})^2+(\sqrt{2})^2}=\sqrt{2+2}=2\\ \\ |b|=\sqrt{(\sqrt{3})^2+1}=\sqrt{3+1}=2](https://tex.z-dn.net/?f=%7Ca%7C%3D%5Csqrt%7B%28%5Csqrt%7B2%7D%29%5E2%2B%28%5Csqrt%7B2%7D%29%5E2%7D%3D%5Csqrt%7B2%2B2%7D%3D2%5C%5C%20%5C%5C%20%7Cb%7C%3D%5Csqrt%7B%28%5Csqrt%7B3%7D%29%5E2%2B1%7D%3D%5Csqrt%7B3%2B1%7D%3D2)
Поскольку в комплексного числа а
, то данный угол "альфа" лежит в четвертой четверти.
![a=2\Bigg(\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2}\Bigg)=2\Bigg(\cos \Big(-\dfrac{\pi}{4}\Big)+i\sin\Big(-\dfrac{\pi}{4}\Big)\Bigg)](https://tex.z-dn.net/?f=a%3D2%5CBigg%28%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D-i%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%5CBigg%29%3D2%5CBigg%28%5Ccos%20%5CBig%28-%5Cdfrac%7B%5Cpi%7D%7B4%7D%5CBig%29%2Bi%5Csin%5CBig%28-%5Cdfrac%7B%5Cpi%7D%7B4%7D%5CBig%29%5CBigg%29)
Аналогично в комплексного числа b угол 'бетта' лежит в четвертой четверти, следовательно:
![b=2\Bigg(\dfrac{\sqrt{3}}{2}-i\dfrac{1}{2}\Bigg)=2\Bigg(\cos\Big(-\dfrac{\pi}{6}\Big)+i\sin\Big(-\dfrac{\pi}{6}\Big)\Bigg)](https://tex.z-dn.net/?f=b%3D2%5CBigg%28%5Cdfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D-i%5Cdfrac%7B1%7D%7B2%7D%5CBigg%29%3D2%5CBigg%28%5Ccos%5CBig%28-%5Cdfrac%7B%5Cpi%7D%7B6%7D%5CBig%29%2Bi%5Csin%5CBig%28-%5Cdfrac%7B%5Cpi%7D%7B6%7D%5CBig%29%5CBigg%29)
Тогда по формуле Муавра
![a^{26}=2^{26}\Bigg(\cos\Big(-\dfrac{\pi}{4}\cdot 26\Big)+i\sin\Big(-\dfrac{\pi}{4}\cdot 26\Big)\Bigg)=2^{26}\Bigg(\cos \dfrac{13\pi}{2}-i\sin\dfrac{13\pi}{2}\Bigg)\\ \\ \\ b^{26}=2^{26}\Bigg(\cos\Big(-\dfrac{\pi}{6}\cdot 26\Big)+i\sin\Big(-\dfrac{\pi}{6}\cdot 26\Big)\Bigg)=2^{26}\Bigg(\cos\dfrac{13\pi}{3}-i\sin\dfrac{13\pi}{3}\Bigg)](https://tex.z-dn.net/?f=a%5E%7B26%7D%3D2%5E%7B26%7D%5CBigg%28%5Ccos%5CBig%28-%5Cdfrac%7B%5Cpi%7D%7B4%7D%5Ccdot%2026%5CBig%29%2Bi%5Csin%5CBig%28-%5Cdfrac%7B%5Cpi%7D%7B4%7D%5Ccdot%2026%5CBig%29%5CBigg%29%3D2%5E%7B26%7D%5CBigg%28%5Ccos%20%5Cdfrac%7B13%5Cpi%7D%7B2%7D-i%5Csin%5Cdfrac%7B13%5Cpi%7D%7B2%7D%5CBigg%29%5C%5C%20%5C%5C%20%5C%5C%20b%5E%7B26%7D%3D2%5E%7B26%7D%5CBigg%28%5Ccos%5CBig%28-%5Cdfrac%7B%5Cpi%7D%7B6%7D%5Ccdot%2026%5CBig%29%2Bi%5Csin%5CBig%28-%5Cdfrac%7B%5Cpi%7D%7B6%7D%5Ccdot%2026%5CBig%29%5CBigg%29%3D2%5E%7B26%7D%5CBigg%28%5Ccos%5Cdfrac%7B13%5Cpi%7D%7B3%7D-i%5Csin%5Cdfrac%7B13%5Cpi%7D%7B3%7D%5CBigg%29)
Окончательно получим:
![\Bigg(\dfrac{a}{b}\Bigg)^{26}=\dfrac{2^{26}\Bigg(\cos \dfrac{13\pi}{2}-i\sin\dfrac{13\pi}{2}\Bigg)}{2^{26}\Bigg(\cos\dfrac{13\pi}{3}-i\sin\dfrac{13\pi}{3}\Bigg)}=\dfrac{0-i\cdot 1}{\dfrac{1}{2}-i\dfrac{\sqrt{3}}{2}}=-\dfrac{2i}{1-i\sqrt{3}}=\\ \\ \\ =-\dfrac{2i(1+i\sqrt{3})}{(1-i\sqrt{3})(1+i\sqrt{3})}=-\dfrac{2i(1+i\sqrt{3})}{1+3}=-\dfrac{i(1+i\sqrt{3})}{2}=\dfrac{\sqrt{3}}{2}-\dfrac{1}{2}i](https://tex.z-dn.net/?f=%5CBigg%28%5Cdfrac%7Ba%7D%7Bb%7D%5CBigg%29%5E%7B26%7D%3D%5Cdfrac%7B2%5E%7B26%7D%5CBigg%28%5Ccos%20%5Cdfrac%7B13%5Cpi%7D%7B2%7D-i%5Csin%5Cdfrac%7B13%5Cpi%7D%7B2%7D%5CBigg%29%7D%7B2%5E%7B26%7D%5CBigg%28%5Ccos%5Cdfrac%7B13%5Cpi%7D%7B3%7D-i%5Csin%5Cdfrac%7B13%5Cpi%7D%7B3%7D%5CBigg%29%7D%3D%5Cdfrac%7B0-i%5Ccdot%201%7D%7B%5Cdfrac%7B1%7D%7B2%7D-i%5Cdfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D%7D%3D-%5Cdfrac%7B2i%7D%7B1-i%5Csqrt%7B3%7D%7D%3D%5C%5C%20%5C%5C%20%5C%5C%20%3D-%5Cdfrac%7B2i%281%2Bi%5Csqrt%7B3%7D%29%7D%7B%281-i%5Csqrt%7B3%7D%29%281%2Bi%5Csqrt%7B3%7D%29%7D%3D-%5Cdfrac%7B2i%281%2Bi%5Csqrt%7B3%7D%29%7D%7B1%2B3%7D%3D-%5Cdfrac%7Bi%281%2Bi%5Csqrt%7B3%7D%29%7D%7B2%7D%3D%5Cdfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D-%5Cdfrac%7B1%7D%7B2%7Di)
Смотри ответ в прилагаемом файле
1. cosx=1 y=-6 cosx=-1 y=7+1=8
y∈(-6;8)
2. y=2x²-x-1 F=∫(2x²-x-1)dx=2x³/3-x²/2-x+C
3=0+C C=3 2x³/3-x²/2-x+3