已知x/y=2/7,求分式x2-3xy+2y2/x2-2xy+y2的值

题目

已知x/y=2/7,求分式x2-3xy+2y2/x2-2xy+y2的值

答案

x2-3xy+2y2/x2-2xy+y2
分子分母同时除以y²得
原式=[(x/y)²-3(x/y)+2]/[(x/y)²-2(x/y)+1]
=(4/49-6/7+2)/(4/49-4/7+1)
=(4-42+98)/(4-28+49)
=60/25
=12/5