若实数x、y、z满足x+1/y=4,y+1/z=1,z+1/x=7/3,则xyz的值为多少
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若实数x、y、z满足x+1/y=4,y+1/z=1,z+1/x=7/3,则xyz的值为多少若实数x、y、z满足x+1/y=4,y+1/z=1,z+1/x=7/3,则xyz的值为多少若实数x、y、z满足
若实数x、y、z满足x+1/y=4,y+1/z=1,z+1/x=7/3,则xyz的值为多少
若实数x、y、z满足x+1/y=4,y+1/z=1,z+1/x=7/3,则xyz的值为多少
若实数x、y、z满足x+1/y=4,y+1/z=1,z+1/x=7/3,则xyz的值为多少
(X+1/Y)(Z+1/X)(Y+1/Z)=XYZ+1/(XYZ)+X+Y+Z+1/X+1/Y+1/Z=XYZ+1/(XYZ)+4+1+7/3=4*1*7/3
得到:
XYZ+1/(XYZ)=2,解得XYZ=1
参考资料里有其他人做的原题
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