一道数学很复杂的题目
发布网友
发布时间：2022-04-24 14:39
共2个回答
热心网友
时间：2023-10-16 20:14
1.为了简化算式，设A=2000的立方根，B=2001的立方根，C=2002的立方根
A^3=2000，B^3=2001，C^3=2002
A^3x^3=B^3y^3=C^z^3
y=Ax/B
z=Ax/C
2000x^2+2002z^2+2001y^2
=A^3*x^2+C^3*(Ax/C)^2+B^3*(Ax/B}^2
=A^3*x^2+A^2*C*x^2+A^2*B*x^2
=A^2*X^2*(A+B+C)
(2000x^2+2002z^2+2001y^2)的立方根=A+B+C
2000x^2+2002z^2+2001y^2=(A+B+C)^3
2000x^2+2002z^2+2001y^2=A^2*X^2*(A+B+C)
A^2*x^2*(A+B+C)=(A+B+C)^3
A^2*x^2=(A+B+C)^2
Ax=A+B+C
1/x=A/(A+B+C)
y=Ax/B=(A+B+C)/B
1/y=B/(A+B+C)
z=Ax/C=(A+B+C)/C
1/z=C/(A+B+C
1/x+1/y+1/z=A/(A+B+C)+B/(A+B+C)+C/(A+B+C=1
热心网友
时间：2023-10-16 20:15
2000x^3=2001y^3=2002z^3=a
则，2000x^2=a/x,2001y^2=a/y,2002z^2=a/z
(2000x^2+2001y^2+2002z^2)^(1/3)=2000^(1/3)+2001^(1/3)+2002^(1/3)=b
(a/x+a/y+a/z)^(1/3)=b
a(1/x+1/y+1/z)=b^3
(1/x+1/y+1/z)=b^3/a
(1/x+1/y+1/z)=(2000^(1/3)/(2000^(1/3)x)+2001^(1/3)/(2001^(1/3)y)+2002^(1/3)/(2002^(1/3)x))^3
(1/x+1/y+1/z)=(1/x+1/y+1/z)^3
假设1/x+1/y+1/z=A
A^3-A=0
A(A-1)(A+1)=0
那么A=0,A=-1，A=1
1/x+1/y+1/z=0，1/x+1/y+1/z=-1，1/x+1/y+1/z=1
