x^2-7xy+12y^2=0,求x^2-2xy+y^2/2xy的值
问题描述：
x^2-7xy+12y^2=0,求x^2-2xy+y^2/2xy的值
答
x^2-7xy+12y^2=0==>(X-3Y)(X-4Y)=0==>X=3Y 或者X=4Y
当X=3Y
x^2-2xy+y^2/2xy=((3Y)^2-2*3Y*Y+Y^2)/(2*3Y*Y)=4Y^2/6Y^2=2/3
当X=4Y
x^2-2xy+y^2/2xy=((4Y)^2-2*4Y*Y+Y^2)/(2*4Y*Y)=9Y^2/8Y^2=9/8
所以x^2-2xy+y^2/2xy值为2/3或者9/8
答
x^2-7xy+12y^2=0(x-3y)(x-4y)=0x=3y或x=4y当x=3y时 (x^2-2xy+y^2)/2xy=(x-y)^2/2xy=(3y-y)^2/2*3y*y=4y^2/6y^2=4/6=2/3当x=4y时 (x^2-2xy+y^2)/2xy=(x-y)^2/2xy=(4y-y)^2/2*4y*y=9y^2/8y^2=9/8