《2025年千里马单元测试卷八年级数学下册人教版》

1. 先化简,再求值:$a(\sqrt{2}-a)+(a + \sqrt{3})(a - \sqrt{3})$,其中$a = \sqrt{2}-1$.

答案： 解:原式$=\sqrt{2}a - a^{2}+a^{2}-3=\sqrt{2}a - 3$. 当$a = \sqrt{2}-1$时, 原式$=\sqrt{2}(\sqrt{2}-1)-3 = 2-\sqrt{2}-3=-1-\sqrt{2}$.

2. 已知$x = \sqrt{6}-\sqrt{5},y = \sqrt{6}+\sqrt{5}$,求$x^{3}y + xy^{3}$的值.

答案： 解:$\because x=\sqrt{6}-\sqrt{5},y=\sqrt{6}+\sqrt{5}$, $\therefore xy = (\sqrt{6}-\sqrt{5})(\sqrt{6}+\sqrt{5}) = 1$, $x + y=\sqrt{6}-\sqrt{5}+\sqrt{6}+\sqrt{5}=2\sqrt{6}$, $\therefore x^{3}y+xy^{3}=xy(x^{2}+y^{2}) = xy[(x + y)^{2}-2xy]=1\times[(2\sqrt{6})^{2}-2\times1]=22$.