初二的一道数学题.(关于实数)

显示全部楼层 · 2008-10-23 18:47:07

若2000X^3=2001Y^3=2002Z^3,XYZ>0,且:
三次根号2000X^2+2001Y^2+2002Z^2
=三次根号2000+三次根号2001+三次根号2002,
求1/X+1/Y+1/Z的值.

千问 · 2008-10-23 18:47:07

令2000x^3=2001y^3=2002z^3=N，题中方程立方得 2000x^2+2001y^2+2000z^2=（3次√2000+3次√2001+3次√2002）^3 两边同除以N 左边：(2000x^2+2001y^2+2000z^2)/N=2000x^2/N+2001y^2/N+2000z^2/N =2000x^2/2000x^3+2001y^2/2001y^3+2000z^2/2002z^3 =1/x+1/y+1/z 右边：（3次√2000+3次√2001+3次√2002）^3/N =[(3次√2000/3次√N)+(3次√2001/3次√N)+(3次√2002/3次√N)]^3 =[(3次√2000/3...