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八年级数学计算题特难
题目

化简求值:已知1x+1y=3\dfrac{1}{x}+\dfrac{1}{y}=3,求2xxy+2yx2xy+y\dfrac{2x-xy+2y}{x-2xy+y}的值

知识点:整体代入法,分式的化简求值章节:未标注

答案与解析

答案

解:1x+1y=3∵ \dfrac{1}{x}+ \dfrac{1}{y}=3

x+y=3xy∴x+y=3xy

2xxy+2yx2xy+y∴ \dfrac{2x-xy+2y}{x-2xy+y}

=2(x+y)xy(x+y)2xy= \dfrac{2\left(x+y\right)-xy}{\left(x+y\right)-2xy}

=6xyxy3xy2xy= \dfrac{6xy-xy}{3xy-2xy}

=5xyxy= \dfrac{5xy}{xy}

=5=5

解析

解:1x+1y=3∵ \dfrac{1}{x}+ \dfrac{1}{y}=3

x+y=3xy∴x+y=3xy

2xxy+2yx2xy+y∴ \dfrac{2x-xy+2y}{x-2xy+y}

=2(x+y)xy(x+y)2xy= \dfrac{2\left(x+y\right)-xy}{\left(x+y\right)-2xy}

=6xyxy3xy2xy= \dfrac{6xy-xy}{3xy-2xy}

=5xyxy= \dfrac{5xy}{xy}

=5=5

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