【几何求解题】难度及重要性属高大上……

风花飘飘

红树

A和G在圆上，求出∠B度数，有点旋…

红树

条件：A和G在圆上，求出∠B度数，有点旋…

红树

条件：A和G在圆上，求出∠B度数，有点旋…

风花飘飘

方程 2*r^2-(64^(1/5)+5*4^(1/5))*r-2*4^(1/5)=0 的解为 r

解 1
r = -(sqrt(25*4^(2/5)+(5*2^(11/5)+16)*4^(1/5)+2^(12/5))-5*4^(1/5)-2^(6/5))/4  

解 2
r = (sqrt(25*4^(2/5)+(5*2^(11/5)+16)*4^(1/5)+2^(12/5))+5*4^(1/5)+2^(6/5))/4


方程 x^2-((sqrt(25*4^(2/5)+(5*2^(11/5)+16)*4^(1/5)+2^(12/5))+5*4^(1/5)+2^(6/5))/4)*x-4^(1/5) 的解为 x

解 1
x = -(sqrt((10*4^(1/5)+2^(11/5))*sqrt(25*4^(2/5)+(5*2^(11/5)+16)*4^(1/5)+2^(12/5))+50*4^(2/5)+(5*2^(16/5)+80)*4^(1/5)+2^(17/5))-sqrt(25*4^(2/5)+(5*2^(11/5)+16)*4^(1/5)+2^(12/5))-5*4^(1/5)-2^(6/5))/8  

解 2
x = (sqrt((10*4^(1/5)+2^(11/5))*sqrt(25*4^(2/5)+(5*2^(11/5)+16)*4^(1/5)+2^(12/5))+50*4^(2/5)+(5*2^(16/5)+80)*4^(1/5)+2^(17/5))+sqrt(25*4^(2/5)+(5*2^(11/5)+16)*4^(1/5)+2^(12/5))+5*4^(1/5)+2^(6/5))/8

风花飘飘

-1024^(1/5)+256^(1/5)*(2*r^2-2*s)-64^(1/5)*(s^2+2*s*r^2)+10*16^(1/5)*r+4^(1/5)*(4*r^3+2*s*r)+4=0

方程 -1024^(1/5)+256^(1/5)*(2*r^2-2*s)-64^(1/5)*(s^2+2*s*r^2)+10*16^(1/5)*r+4^(1/5)*(4*r^3+2*s*r)+4=0 的解为 s
解 1
s = -(sqrt(2^(22/5)*r^4+2^(21/5)*4^(1/5)*r^3+(2^(31/5)*8^(1/5)+4^(7/5))*r^2+(5*2^(21/5)*16^(1/5)-4^(11/5)*8^(1/5))*r+2*8^(7/5))+2^(11/5)*r^2-2*4^(1/5)*r+4*8^(1/5))/2^(11/5)  

解 2
s = (sqrt(2^(22/5)*r^4+2^(21/5)*4^(1/5)*r^3+(2^(31/5)*8^(1/5)+4^(7/5))*r^2+(5*2^(21/5)*16^(1/5)-4^(11/5)*8^(1/5))*r+2*8^(7/5))-2^(11/5)*r^2+2*4^(1/5)*r-4*8^(1/5))/2^(11/5)
= (sqrt(2^(12/5)*r^4+2^(13/5)*r^3+17*2^(4/5)*r^2+32*r+2^(16/5))-2^(6/5)*r^2+2^(2/5)*r-2^(8/5))/2^(6/5)

方程 -1024^(1/5)+256^(1/5)*(2*r^2-2*s)-64^(1/5)*(s^2+2*s*r^2)+10*16^(1/5)*r+4^(1/5)*(4*r^3+2*s*r)+4=0,s=0 的解为 r,s
解 1
r = -(4^(4/5)*sqrt(8^(2/5)-10*4^(1/5)*16^(1/5))+4^(4/5)*8^(1/5))/8,s = 0  

解 2
r = (4^(4/5)*sqrt(8^(2/5)-10*4^(1/5)*16^(1/5))-4^(4/5)*8^(1/5))/8,s = 0

解 3
r = 0,s = 0


程 x^2-((4^(4/5)*sqrt(8^(2/5)-10*4^(1/5)*16^(1/5))-4^(4/5)*8^(1/5))/8)*x-4^(1/5)=0 的解为 x
解 1
x = -(4^(4/5)*sqrt(-2*8^(1/5)*sqrt(8^(2/5)-10*4^(1/5)*16^(1/5))-10*4^(1/5)*16^(1/5)+2*8^(2/5)+4^(13/5))-4^(4/5)*sqrt(8^(2/5)-10*4^(1/5)*16^(1/5))+4^(4/5)*8^(1/5))/16

解 2
x = (4^(4/5)*sqrt(-2*8^(1/5)*sqrt(8^(2/5)-10*4^(1/5)*16^(1/5))-10*4^(1/5)*16^(1/5)+2*8^(2/5)+4^(13/5))+4^(4/5)*sqrt(8^(2/5)-10*4^(1/5)*16^(1/5))-4^(4/5)*8^(1/5))/16





表达式 (x^3+r*x^2+s*x+8^(1/5))*(x^2-r*x-4^(1/5)) = x^5+(s-r^2-2^(2/5))*x^3+(-r*s-2^(2/5)*r+2^(3/5))*x^2+(-2^(2/5)*s-2^(3/5)*r)*x-2










x^3+r*x^2+((sqrt(2^(12/5)*r^4+2^(13/5)*r^3+17*2^(4/5)*r^2+32*r+2^(16/5))-2^(6/5)*r^2+2^(2/5)*r-2^(8/5))/2^(6/5))*x+8^(1/5)=0
x^2-r*x-4^(1/5)=0

所以：
(x^3+r*x^2+((sqrt(2^(12/5)*r^4+2^(13/5)*r^3+17*2^(4/5)*r^2+32*r+2^(16/5))-2^(6/5)*r^2+2^(2/5)*r-2^(8/5))/2^(6/5))*x+8^(1/5))*(x^2-r*x-4^(1/5))=0
= (4*x^5+sqrt(2^(12/5)*r^4+2^(13/5)*r^3+17*2^(4/5)*r^2+32*r+2^(16/5))*(2^(4/5)*x^3-2^(4/5)*r*x^2-2^(6/5)*x)+(-8*r^2+2^(6/5)*r-2^(17/5))*x^3+(4*r^3-2^(6/5)*r^2+2^(13/5))*x^2+(2^(12/5)*r^2-3*2^(8/5)*r+2^(14/5))*x-8)/4

suoyi:
sqrt(2^(12/5)*r^4+2^(13/5)*r^3+17*2^(4/5)*r^2+32*r+2^(16/5))*(2^(4/5)*x^3-2^(4/5)*r*x^2-2^(6/5)*x)+(-8*r^2+2^(6/5)*r-2^(17/5))*x^3+(4*r^3-2^(6/5)*r^2+2^(13/5))*x^2+(2^(12/5)*r^2-3*2^(8/5)*r+2^(14/5))*x+5*x=0

即
方程 sqrt(2^(12/5)*r^4+2^(13/5)*r^3+17*2^(4/5)*r^2+32*r+2^(16/5))*(2^(4/5)*x^2-2^(4/5)*r*x-2^(6/5))+(-8*r^2+2^(6/5)*r-2^(17/5))*x^2+(4*r^3-2^(6/5)*r^2+2^(13/5))*x+2^(12/5)*r^2-3*2^(8/5)*r+2^(14/5)+5=0 的解为 r

ccmmjj

是个好题目。要求一个四次方程的解，这个角不是特殊角。