设x=【1/3】y,求【3y/x+3y】+【x/3y-x】-【6xy/9y²-x²】的值.

seshot2022-10-04 11:39:541条回答

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zhangping444 共回答了16个问题 | 采纳率75%: 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; 1年前

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