11. （1）已知$x = \sqrt{2} + 1$，求代数式$x^2 - 2x + 2$的值；
（2）（2023·淄博改编）已知$x = \frac{\sqrt{5} - 1}{2}$，$y = \frac{\sqrt{5} + 1}{2}$，求$x^2 + xy + y^2$的值.
（2）（2023·淄博改编）已知$x = \frac{\sqrt{5} - 1}{2}$，$y = \frac{\sqrt{5} + 1}{2}$，求$x^2 + xy + y^2$的值.
答案
(1) $\because x = \sqrt{2}+1$，$\therefore x - 1 = \sqrt{2}$. $\therefore (x - 1)^2 = 2$，即$x^2 - 2x + 1 = 2$. $\therefore x^2 - 2x = 1$. $\therefore x^2 - 2x + 2 = 1 + 2 = 3$
(2) $\because x = \frac{\sqrt{5}-1}{2}$，$y = \frac{\sqrt{5}+1}{2}$，$\therefore x + y = \frac{\sqrt{5}-1}{2}+\frac{\sqrt{5}+1}{2}=\sqrt{5}$，$xy = \frac{\sqrt{5}-1}{2}\times\frac{\sqrt{5}+1}{2}=1$. $\therefore x^2 + xy + y^2 = x^2 + 2xy + y^2 - xy = (x + y)^2 - xy = (\sqrt{5})^2 - 1 = 4$
(2) $\because x = \frac{\sqrt{5}-1}{2}$，$y = \frac{\sqrt{5}+1}{2}$，$\therefore x + y = \frac{\sqrt{5}-1}{2}+\frac{\sqrt{5}+1}{2}=\sqrt{5}$，$xy = \frac{\sqrt{5}-1}{2}\times\frac{\sqrt{5}+1}{2}=1$. $\therefore x^2 + xy + y^2 = x^2 + 2xy + y^2 - xy = (x + y)^2 - xy = (\sqrt{5})^2 - 1 = 4$
12. 设$6 - \sqrt{10}$的整数部分为$a$，小数部分为$b$，求$(2a + \sqrt{10})b$的值.
答案
$\because 3 ＜ \sqrt{10}＜4$，$\therefore -4 ＜ -\sqrt{10}＜-3$. $\therefore 2 ＜ 6 - \sqrt{10}＜3$. $\therefore 6 - \sqrt{10}$的整数部分$a = 2$，小数部分$b = 6 - \sqrt{10}-2 = 4 - \sqrt{10}$. $\therefore (2a+\sqrt{10})b=(2\times2+\sqrt{10})\times(4 - \sqrt{10})=(4+\sqrt{10})\times(4 - \sqrt{10}) = 4^2 - (\sqrt{10})^2 = 6$
13. （2023·苏州）如图，在平面直角坐标系中，点$A$的坐标为(9,0)，点$C$的坐标为(0,3)，以$OA$、$OC$为邻边作矩形$OABC$. 动点$E$、$F$分别从点$O$、$B$同时出发，以每秒1个单位长度的速度沿$OA$、$BC$向终点$A$、$C$移动，连接$AC$、$EF$. 当移动时间为4秒时，$AC \cdot EF$的值为________.
答案
30 解析：$\because$四边形$OABC$为矩形，$A(9,0)$，$C(0,3)$，$\therefore B(9,3)$，$OC = 3$，$OA = 9$. $\therefore AC=\sqrt{OC^2 + OA^2}=\sqrt{3^2 + 9^2}=3\sqrt{10}$. $\because OE = BF = 4$，$\therefore E(4,0)$，$F(5,3)$. $\therefore$易得$EF=\sqrt{10}$. $\therefore AC\cdot EF = 3\sqrt{10}\times\sqrt{10}=30$.
