1道数学,急!!!
发布网友
发布时间：2024-01-04 02:23
共2个回答
热心网友
时间：2024-07-13 20:10
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：2024-07-13 20:10
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次√2000x^3)+(3次√2001/3次√2001y^3)+(3次√2002/3次√2002z^3)]
=(1/x+1/y+1/z)^3
即是 1/x+1/y+1/z=(1/x+1/y+1/z)^3
化简 (1/x+1/y+1/z)^2=1
因为 2000x^3=2001y^3=2002z^3且xyz大于0
所以 x>0,y>0,z>0
所以 1/x+1/y+1/z=1
