设x>3y>0,2log3(x-3y)=log3(3x+y)+log3y,求x/y的值
人气:317 ℃ 时间:2020-09-07 21:17:08
解答
2log3(x-3y)=log3(3x+y)+log3ylog3(x-3y)^2=log3[y(3x+y)](x-3y)^2=y(3x+y)x^2-6xy+9y^2=3xy+y^2x^2-9xy+8y^2=0(x-y)(x-8y)=0x=y,x=8yx>3y>0则x=y显然不成立所以x=8yx/y=8
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