22. (本题9分)根据所学知识,我们通过证明可以得到一个定理:一个非零有理数与一个无理数的积仍为一个无理数,根据这个定理得到一个结论:若$x+y\sqrt{m}=0$,其中$x,y$是有理数,$\sqrt{m}$是无理数,则$x=0,y=0$.证明:$\because x+y\sqrt{m}=0,x$是有理数,$\therefore y\sqrt{m}$是有理数.$\because y$是有理数,$\sqrt{m}$是无理数,$\therefore y=0,\therefore x+0×\sqrt{m}=0,\therefore x=0$.
(1)若$x+\sqrt{2}y=\sqrt{2}(1-\sqrt{2})$,其中$x,y$是有理数,则$x=$,$y=$;
(2)若$x+y\sqrt{m}=a+b\sqrt{m}$,其中$x,y,a,b$是有理数,$\sqrt{m}$是无理数,求证:$x=a,y=b$;
(3)已知$\sqrt{17}$的整数部分为$a$,小数部分为$b,x,y$是有理数,$x,y,a,b$满足$17y+\sqrt{17}y+\sqrt{17}(y-2\sqrt{17}x)=2a\sqrt{17}+b\sqrt{17}$,求$x,y$的值.

22.【解析】(1)$-2$ $1$
(2)证明:$\because x+y\sqrt{m}=a+b\sqrt{m}$,
$\therefore x-a+(y-b)\sqrt{m}=0$,
$\because x,y,a,b$为有理数,
$\therefore x-a,y-b$都是有理数,
$\therefore x-a=0$,$y-b=0$,$\therefore x=a$,$y=b$.
(3)$\because 4＜\sqrt{17}＜5$,又知$\sqrt{17}$的整数部分为$a$,小数部分为$b$,$\therefore a=4$,$b=\sqrt{17}-4$,
$\because 17y+\sqrt{17}y+\sqrt{17}(y-2\sqrt{17}x)=2a\sqrt{17}+b\sqrt{17}$,
$\therefore 17y+\sqrt{17}y+\sqrt{17}y-34x=8\sqrt{17}+\sqrt{17}(\sqrt{17}-4)$,
即$17y-34x+2\sqrt{17}y=17+4\sqrt{17}$,
$\because x,y$是有理数,$\therefore \begin{cases}17y-34x=17,\\2y=4,\end{cases}$解得$\begin{cases}x=\dfrac{1}{2},\\y=2.\end{cases}$

23.(本题12分)如图是一种躺椅及其简化结构示意图,扶手AB与底座CD都平行于地面EF,前支架OE与后支架OF分别与CD交于点G和点D,AB与DM相交于点N,∠AOE=∠BNM.
(1)求证:OE//DM;
(2)若OE平分∠AOF,∠ODC=30°,求扶手AB与靠背DM的夹角∠ANM的度数.

23.【解析】(1)证明:$\because ∠BNM=∠AND$,$∠AOE=∠BNM$,$\therefore ∠AOE=∠AND$,$\therefore OE// DM.$
(2)$\because AB$与$CD$都平行于$EF$,
$\therefore AB// CD$,$\therefore ∠BOD=∠ODC=30°.$
$\because ∠AOF+∠BOD=180°$,$\therefore ∠AOF=150°.$
$\because OE$平分$∠AOF$,$\therefore ∠EOF=\dfrac{1}{2}∠AOF=75°.$
$\therefore ∠BOE=∠BOD+∠EOF=105°.$
$\because OE// DM$,$\therefore ∠ANM=∠BOE=105°.$