数学，求步骤，急救，第一题
2个回答
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解原式
=(logxy+logx^(1/2)y^(3/2))/(logx^(3/4)+logy^2)
=(logx^(3/2)y^(5/2))/(logx^(3/4)y^2)
=log(x^(3/2)y^(5/2))/x^(3/4)y^2)
=log(x^(3/2-3/4)y^(5/2-2))
=logx^(3/4)y^(1/2)
=1/2logx^(3/2)y
=(logxy+logx^(1/2)y^(3/2))/(logx^(3/4)+logy^2)
=(logx^(3/2)y^(5/2))/(logx^(3/4)y^2)
=log(x^(3/2)y^(5/2))/x^(3/4)y^2)
=log(x^(3/2-3/4)y^(5/2-2))
=logx^(3/4)y^(1/2)
=1/2logx^(3/2)y
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