Question

高一数学题目吖。

已知x、y、z均为正实数，且3的x次方=4的y次方=6的z次方
（1）求证 1/z - 1/x= 1/2y
（2）试比较3x、4y、6z的大小

Answer 1

()前面的数是底数,()里的数是真数
1.令3^x=4^y=6^z=k，k＞1
x=log3(k),y=log4(k),z=log6(k)
1/x=logk(3),1/y=logk(4),1/z=logk(6)
1/z - 1/x=logk(6)-logk(3)=logk(6/3)=logk(2),
1/2y=(1/2)logk(4)=logk(2)
所以1/z - 1/x= 1/2y
2.
3x=3log3(k)
=(1/4)*12*log3(k)
=(1/4)log3(k^12)
=log3^4(k^12)
=1/[logk^12(3^4)]
=1/[logk^12(81)]

4y=4log4(k)
=(1/3)*12*log4(k)
=(1/3)log4(k^12)
=log4^3(k^12)
=1/[logk^12(4^3)]
=1/[logk^12(64)]

6z=6log6(k)
=(1/2)*12*log6(k)
=(1/2)log6(k^12)
=log6^2(k^12)
=1/[logk^12(6^2)]
=1/[logk^12(36)]

k^12 > 1
0 < logk^12(36) < logk^12(64) < logk^12(81)
1/[logk^12(36)] > 1/[logk^12(64)] > 1/[logk^12(81)]
所以3x<4y<6z

一定采纳啊，写了好久。