已知x/y=2/7 则x^2-3xy+2y^2 / 2x^2-3xy+7y^2=?1/x+1/y=2 求2x-3xy+2y / 3x+3y+2xy=?
问题描述：
已知x/y=2/7 则x^2-3xy+2y^2 / 2x^2-3xy+7y^2=?1/x+1/y=2 求2x-3xy+2y / 3x+3y+2xy=?
已知x/y=2/7 则x^2-3xy+2y^2 / 2x^2-3xy+7y^2=?
1/x+1/y=2 求2x-3xy+2y / 3x+3y+2xy=?
答
1,(x^2-3xy+2y^2) / (2x^2-3xy+7y^2)
=(x-2y)(x-y)/[(2x-y)(x-y)+6y^2],
由x/y=2/7,得：7x=2y,
所以x-y=-5y/7,x-2y=-12y/7,2x-y=-3y/7,
所以原式=[(-12y/7)(-5y/7)]/[(-3y/7)(-5y/7)+6y^2]
=(60/49)/[(15/49)+6]
=60/309=20/103.
或(x^2-3xy+2y^2) / (2x^2-3xy+7y^2) （分子,分母同除以y^2）
=[(x/y)^2-3(x/y)+2]/[2(x/y)^2-3(x/y)+7]
=[(2/7)^2-3*2/7+2]/[2*(2/7)^2-3*2/7+7]
=[4/49-6/7+2]/[8/49-6/7+7]
=(60/49)/(309/49)=20/103.
2,由1/x+1/y=2,得：x+y=2xy,
所以(2x-3xy+2y) / (3x+3y+2xy)
=[2(x+y)-3xy]/[3(x+y)+2xy]
=[4xy-3xy]/[6xy+2xy]
=xy/(8xy)=1/8.