题目内容
11.先化简,后求值:x2+y2-2x+2y+2,其中x=$\sqrt{2}$+1,y=$\sqrt{2}$-1.分析 先将x2+y2-2x+2y+2变形为(x+y)2-2xy-2(x-y)+2,然后将x和y的值代入求解即可.
解答 解:∵x=$\sqrt{2}$+1,y=$\sqrt{2}$-1,
∴x+y=2$\sqrt{2}$,x-y=2,xy=2-1=1,
∴x2+y2-2x+2y+2
=(x+y)2-2xy-2(x-y)+2
=(2$\sqrt{2}$)2-2×1-2×2+2
=8-2-4+2
=4.
点评 本题考查了二次根式的化简,解答本题的关键在于先将x2+y2-2x+2y+2变形为(x+y)2-2xy-2(x-y)+2,然后将x和y的值代入求解.
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