已知X/Y=2/3,则X^2-Y^2/X^2-2XY+Y^2除以XY+Y^2/X^2-XY

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已知X/Y=2/3,则X^2-Y^2/X^2-2XY+Y^2除以XY+Y^2/X^2-XY已知X/Y=2/3,则X^2-Y^2/X^2-2XY+Y^2除以XY+Y^2/X^2-XY已知X/Y=2/3,

已知X/Y=2/3,则X^2-Y^2/X^2-2XY+Y^2除以XY+Y^2/X^2-XY
已知X/Y=2/3,则X^2-Y^2/X^2-2XY+Y^2除以XY+Y^2/X^2-XY

已知X/Y=2/3,则X^2-Y^2/X^2-2XY+Y^2除以XY+Y^2/X^2-XY
原式=(X+Y)(X-Y)/(X-Y)^2÷Y(X+Y)/(X+Y)(X-Y)
=(X+Y)/(X-Y)÷Y(X-Y)
=(X+Y)/(X-Y)×(X-Y)/Y
=(X+Y)/Y
=X/Y+Y/Y
=2/3+1
=5/3

根号下大于等于0
1/x≥0
x>0
真数大于0
x>0
分母不等于0
1-lgx≠0
lgx≠1
x≠10
所以定义域(0,10)∪(10,+∞)

是不是没有加括号啊；如果是（X^2-Y^2）/（X^2-2XY+Y^2）除以（XY+Y^2）/（X^2-XY）的话，分子分母分别进行化简就得到X/Y,很显然等于2/3啊

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