Question

一道简单的高中数学题，哪位进来看下帮帮忙

已知实数a，b满足a-3a^(2/3)+5a^(1/3)=1, b-3b^(2/3)+5b^(1/3)=5,求a^(1/3)+b^(1/3)

Answer 1

令a^(1/3)=x, b^(1/3)=y 化简
a-3a^(2/3)+5a^(1/3)=1
x^3-3x^2+5x-1=0
x(x-1)(x-2)+3(x-1)=-2
(x-1)((x-1)^2+2)=-2
(x-1)^3+2(x-1)=-2
b-3b^(2/3)+5b^(1/3)=5
y^3-3y^2+5y-5=0
y(y-1)(y-2)+3(y-1)=2
(y-1)^3+2(y-1)=2
函数f(x')=x'^3+2x'=x'(x'^2+2)
f(-x')=-f(x')
f(x')为原点对称的奇函数
上述方程组可看成 f(x')=2 f(x')=-2
即 y-1=x0
x-1=-x0
x-1+y-1=x0+(-x0)=0
即x+y=2

Answer 2

因为a-3a^(2/3)+5a^(1/3)=1
令a^(1/3)=T
得：T^3-3T^2+T=1
求得T=2
又因为b-3b^(2/3)+5b^(1/3)=5
令b^(1/3)=F
得F^3-3F^2+5F=5
求得：F
再求T+F= 就是答案喇