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设x=【1/3】y,求【3y/x+3y】+【x/3y-x】-【6xy/9y²-x²】的值.
人气:466 ℃ 时间:2019-12-19 02:04:15
解答
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
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