∵lgx+lgy=2lg(x-2y)
∴lg(xy)=lg(x-2y)^2
∴xy=(x-2y)^2
xy=(x-2y)^2
xy=x^2-4xy+4y^2
同时除以y^2
(x/y)^2-5(x/y)+4=0
将“x/y”换为“t”
t^2-5t+4=0
配方
t^2-5t+(5/2)^2=9/4
t-5/2=±3/2
t1=1 t2=4
∵(x/y)=t
∴(x/y)=1或4
则log[2^(1/2)][(x/y)]为0或4
已知:lgx+lgy=2lg(x-2y),则log[2^(1/2)][(x/y)]的值为____
已知:lgx+lgy=2lg(x-2y),则log[2^(1/2)][(x/y)]的值为____
数学人气:813 ℃时间:2020-05-20 06:29:29
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