《2025年暑假作业新疆青少年出版社八年级数学人教版》

10.已知$a^{2}+b^{2}-4a-2b+5= 0$,求$\frac {\sqrt {a}+b}{3b-2\sqrt {a}}$的值.

答案： $\because a^{2}+b^{2}-4a-2b+5=0,\therefore (a-2)^{2}+(b-1)^{2}=0,\therefore a=2,b=1,\therefore \frac {\sqrt {a}+b}{3b-2\sqrt {a}}=\frac {\sqrt {2}+1}{3-2\sqrt {2}}=7+5\sqrt {2}$

11.已知$x= \frac {1}{2+\sqrt {3}},y= \frac {1}{2-\sqrt {3}}$,求代数式$x^{2}-2xy+y^{2}$的值.

答案： $\because x=\frac {1}{2+\sqrt {3}}=2-\sqrt {3},y=\frac {1}{2-\sqrt {3}}=2+\sqrt {3},\therefore x^{2}-2xy+y^{2}=(x-y)^{2}=[(2-\sqrt {3})-(2+\sqrt {3})]^{2}=(-2\sqrt {3})^{2}=12$

12.已知$\sqrt {x}= \frac {1}{2-\sqrt {3}},\sqrt {y}= \frac {1}{2+\sqrt {3}}$,求$\frac {x-y}{\sqrt {x}-\sqrt {y}}$的值.

答案： 原式$=\frac {(\sqrt {x})^{2}-(\sqrt {y})^{2}}{\sqrt {x}-\sqrt {y}}=\frac {(\sqrt {x}+\sqrt {y})(\sqrt {x}-\sqrt {y})}{\sqrt {x}-\sqrt {y}}=\sqrt {x}+\sqrt {y},\because \sqrt {x}=\frac {1}{2-\sqrt {3}}=2+\sqrt {3},\sqrt {y}=\frac {1}{2+\sqrt {3}}=2-\sqrt {3},\therefore 原式=2+\sqrt {3}+2-\sqrt {3}=4$

13.已知$x= \sqrt {5}+3$,求代数式$x^{3}-x^{2}-26x+5$的值.

答案： $\because x=\sqrt {5}+3,\therefore x^{3}-x^{2}-26x+5=x^{3}-x^{2}-26x+26-26+5=x^{2}(x-1)-26(x-1)-21=(x-1)(x^{2}-26)-21=(\sqrt {5}+2)[(\sqrt {5}+3)^{2}-26]-21=(\sqrt {5}+2)(6\sqrt {5}-12)-21=6-21=-15$