《2025年暑假作业新疆青少年出版社八年级数学人教版》

12. 若等腰三角形两边的长分别是$2\sqrt {3},3\sqrt {2}$,则这个三角形的周长是____.

答案： $6\sqrt{2}+2\sqrt{3}$或$4\sqrt{3}+3\sqrt{2}$

13. 计算:
(1) $9\sqrt {3}+5\sqrt {12}-3\sqrt {48}$;
(2) $3\sqrt {3}-\sqrt {8}+\sqrt {2}-\sqrt {27}$;
(3) $2\sqrt {12}÷\frac {1}{2}\sqrt {50}×2\sqrt {\frac {3}{4}}$;
(4) $(\sqrt {5}+\sqrt {6})^{2024}(\sqrt {5}-\sqrt {6})^{2025}$.

答案：
(1) $7\sqrt{3}$
(2) $-\sqrt{2}$
(3) $\frac{12\sqrt{2}}{5}$
(4) $\sqrt{5}-\sqrt{6}$

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

答案： 原式$=\frac{a - 1 + 1}{a - 1}\div\frac{a(a - 1) - a}{a - 1}=\frac{a}{a - 1}\div\frac{a(a - 2)}{a - 1}=\frac{a}{a - 1}\cdot\frac{a - 1}{a(a - 2)}=\frac{1}{a - 2}$. 当$a=\sqrt{2}+2$时, 原式$=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}$

15. 已知$x= \frac {1}{2}(\sqrt {7}+\sqrt {5}),y= \frac {1}{2}(\sqrt {7}-\sqrt {5})$,求下列各式的值:
(1) $x^{2}+y^{2}$;
(2) $x^{2}-y^{2}$;
(3) $\frac {x}{y}+\frac {y}{x}$.

答案： $\because x=\frac{1}{2}(\sqrt{7}+\sqrt{5}),y=\frac{1}{2}(\sqrt{7}-\sqrt{5}),\therefore x + y=\sqrt{7},x - y=\sqrt{5},xy=\frac{1}{2}$.
(1) $x^{2}+y^{2}=(x + y)^{2}-2xy=(\sqrt{7})^{2}-2\times\frac{1}{2}=6$
(2) $x^{2}-y^{2}=(x + y)(x - y)=\sqrt{7}\times\sqrt{5}=\sqrt{35}$
(3) $\frac{x}{y}+\frac{y}{x}=\frac{x^{2}+y^{2}}{xy}=\frac{6}{\frac{1}{2}}=12$