设x=【1/3】y,求【3y/x+3y】+【x/3y-x】-【6xy/9y²-x²】的值.
问题描述：
设x=【1/3】y,求【3y/x+3y】+【x/3y-x】-【6xy/9y²-x²】的值.
答
3y/(x+3y)+x/(3y-x)-6xy/(9y²-x²)
=(9y²-3xy)/(9y²-x²)+(x²+3xy)/(9y²-x²)-6xy/(9y²-x²)
=(3y-x)²/(9y²-x²)
=(3y-x)/(3y+x)
=[3-(x/y)]/[3+(x/y)]
=[3-(1/3)]/[3+(1/3)]
=4/5