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八年级数学解答题一般
题目
已知:x=12x=1-\sqrt{2},y=1+2y=1+\sqrt{2},求x2+y2xy2x+2yx^{2}+y^{2}-xy-2x+2y的值.
知识点:因式分解的应用、二次根式的性质与化简、二次根式的化简求值章节:第2章 实数 / 2.3 二次根式

答案与解析

答案

x=12\because x=1-\sqrt{2}y=1+2y=1+\sqrt{2}
xy=(12)(1+2)=22\therefore x-y=(1-\sqrt{2})-(1+\sqrt{2})=-2\sqrt{2}
xy=(12)(1+2)=1xy=(1-\sqrt{2})(1+\sqrt{2})=-1
x2+y2xy2x+2y=(xy)22(xy)+xy\therefore x^{2}+y^{2}-xy-2x+2y=\left(x-y\right)^{2}-2\left(x-y\right)+xy
=(22)22×(22)+(1)=(-2\sqrt{2})^{2}-2\times (-2\sqrt{2})+\left(-1\right)
=7+42=7+4\sqrt{2}.

解析

x=12\because x=1-\sqrt{2}y=1+2y=1+\sqrt{2}
xy=(12)(1+2)=22\therefore x-y=(1-\sqrt{2})-(1+\sqrt{2})=-2\sqrt{2}
xy=(12)(1+2)=1xy=(1-\sqrt{2})(1+\sqrt{2})=-1
x2+y2xy2x+2y=(xy)22(xy)+xy\therefore x^{2}+y^{2}-xy-2x+2y=\left(x-y\right)^{2}-2\left(x-y\right)+xy
=(22)22×(22)+(1)=(-2\sqrt{2})^{2}-2\times (-2\sqrt{2})+\left(-1\right)
=7+42=7+4\sqrt{2}.

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