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.

(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) $\because x=\frac{\sqrt{5}-1}{2},y=\frac{\sqrt{5}+1}{2},\therefore x + y=\frac{\sqrt{5}-1}{2}+\frac{\sqrt{5}+1}{2}=\sqrt{5},xy=\frac{\sqrt{5}-1}{2}\times\frac{\sqrt{5}+1}{2}=1.\therefore x^2 + xy + y^2=x^2 + 2xy + y^2-xy=(x + y)^2-xy=(\sqrt{5})^2-1 = 4$

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.

$\because 3＜\sqrt{10}＜4,\therefore -4＜-\sqrt{10}＜-3.\therefore 2＜6-\sqrt{10}＜3.\therefore 6-\sqrt{10}$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________.

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$EF=\sqrt{10}.\therefore AC\cdot EF=3\sqrt{10}\times\sqrt{10}=30$