有理数指数幂 已知 a^(2/3)+b^(2/3)=4,x=a+3*a^(1/3)*b^(2/3),y= b+3*a^(2/3)*b^(1/3),求证:(x+y)^(2/3)+(x-y)^(2/3)为定值.
问题描述：
有理数指数幂 已知 a^(2/3)+b^(2/3)=4,x=a+3*a^(1/3)*b^(2/3),y= b+3*a^(2/3)*b^(1/3),求证:(x+y)^(2/3)+(x-y)^(2/3)为定值.
答
x+y=(a^1/3+b^1/3)^3
x-y=(a^1/3-b^1/3)^3
所以(x+y)^(2/3)+(x-y)^(2/3)为
[(a^1/3+b^1/3)^3]^2/3+[(a^1/3-b^1/3)^3]^2/3
即[a^(1/3)+b^（1/3)]^2+[a^(1/3)-b^（1/3)]^2
=8
ok!
答
x+y=(a^1/3+b^1/3)^3
x-y=(a^1/3-b^1/3)^3
所以(x+y)^(2/3)+(x-y)^(2/3)为
[(a^1/3+b^1/3)^3]^2/3+[(a^1/3-b^1/3)^3]^2/3
即[a^(1/3)+b^（1/3)]^2+[a^(1/3)-b^（1/3)]^2
=8
答
x+y=(a^1/3+b^1/3)^3x-y=(a^1/3-b^1/3)^3所以(x+y)^(2/3)+(x-y)^(2/3)为[(a^1/3+b^1/3)^3]^2/3+[(a^1/3-b^1/3)^3]^2/3即为[a^(1/3)+b^（1/3)]^2+[a^(1/3)-b^（1/3)]^2即为8,是定值我打字好辛苦!