解答题求助1
若2x-3y=0,求(1/2)*xy÷(x^2+y^2)-y^2/(2x^2-y^2)的值
参考答案:由2x-3y=0可知:x=3y/2
将x=3y/2代入
1/2 xy / (x^2+y^2) - y^2 /(2x^2-y^2)
=(1/2×3y/2×y)/[(3y/2)^2+y^2]-y^2/[2×(3y/2)^2-y^2]
=(3y^2/4)/(13y^2/4)-y^2/(7y^2/2)
=3/13-2/7
=21/91-26/91
=-5/91.
若2x-3y=0,求(1/2)*xy÷(x^2+y^2)-y^2/(2x^2-y^2)的值
参考答案:由2x-3y=0可知:x=3y/2
将x=3y/2代入
1/2 xy / (x^2+y^2) - y^2 /(2x^2-y^2)
=(1/2×3y/2×y)/[(3y/2)^2+y^2]-y^2/[2×(3y/2)^2-y^2]
=(3y^2/4)/(13y^2/4)-y^2/(7y^2/2)
=3/13-2/7
=21/91-26/91
=-5/91.