已知P=(x-y分之x^2)-(x-y分之y^2), Q=x-y+（x+y分之2y^2）,当x＞y＞0,比较P,Q大小.O(∩_∩_数学解答

已知P=(x-y分之x^2)-(x-y分之y^2), Q=x-y+（x+y分之2y^2）,当x＞y＞0,比较P,Q大小.O(∩_∩)O谢谢!

人气：253 ℃　时间：2020-06-03 18:31:38

解答

P = x^2/(x-y) -y^2.(x-y) = (x^2-y^2)/(x-y) = x+yQ = x-y+ 2y^2/(x+y) = (x^2 - y^2 +2y^2)/(x+y) = (x^2 + y^2 +2xy -2xy)/(x+y)= x+y - 2xy/(x+y)= P -2xy/(x+y)所以,P>Q.