已知x-3y=0，求x平方-2xy+y平方分之2x+y除以x-y分之1的值
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x-3y=0
x=3y
(2x+y)/(x^2-2xy+y^2)÷1/(x-y)
=(2x+y)/(x-y)^2÷1/(x-y)
=(2x+y)/(x-y)^2*(x-y)
=(2x+y)/(x-y)
=(2*3y+y)/(3y-y)
=7y/2y
=7/2
x=3y
(2x+y)/(x^2-2xy+y^2)÷1/(x-y)
=(2x+y)/(x-y)^2÷1/(x-y)
=(2x+y)/(x-y)^2*(x-y)
=(2x+y)/(x-y)
=(2*3y+y)/(3y-y)
=7y/2y
=7/2
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