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九年级数学解答题一般
题目

已知:x=3+1x=\sqrt{3}+1,y=31y=\sqrt{3}-1,求x22xy+y2x2y2\dfrac{x^{2}-2xy+y^{2}}{x^{2}-y^{2}}的值.

知识点:二次根式的性质与化简章节:未标注

答案与解析

答案

x22xy+y2x2y2\dfrac{x^{2}-2xy+y^{2}}{x^{2}-y^{2}}

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

=xyx+y=\dfrac{x-y}{x+y}(4\ldots (4分)

x=3+1x=\sqrt{3}+1y=31y=\sqrt{3}-1时,原式=223=13=33=\dfrac{2}{2\sqrt{3}}=\dfrac{1}{\sqrt{3}}=\dfrac{\sqrt{3}}{3}.

解析

x22xy+y2x2y2\dfrac{x^{2}-2xy+y^{2}}{x^{2}-y^{2}}

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

=xyx+y=\dfrac{x-y}{x+y}(4\ldots (4分)

x=3+1x=\sqrt{3}+1y=31y=\sqrt{3}-1时,原式=223=13=33=\dfrac{2}{2\sqrt{3}}=\dfrac{1}{\sqrt{3}}=\dfrac{\sqrt{3}}{3}.

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