如图,一次函数$y=x$与反比例函数$y=frac{1}{x}(x>0)$的图象交于点$A$,过点$A$作$ABperpOA$,交$x$轴于点$B;$作$BA_{1}//OA$,交反比例函数图象于点$A_{1}$;过点$A_{1}$作$A_{1}B_{1}perpA_{1}B$交$x$轴于点$B_{1}$;再作$B_{1}A_{2}$$//BA_{1}$,交反比例函数图象于点$A_{2}$,依次进行下去$cdots$,则点$A_{2023}$的横坐标为$_sqrt{2024}+sqrt{2023}_$.<ImageHere>
根据直$OA$的关系式为$y=x$,以及$OAperpAB$,可得到$triangleAOB$是等腰直角三角形,进而得到$triangleA_{2}B_{1}B_{2}、triangleA_{3}B_{2}B_{3}cdotscdots$都是等腰直角三角形,设$OC=a=AC$,则点$A(a,a)$,点$A$在反比例函数..的图象上,可求出$a=1$,进而得到点$A$的横坐标为1,同理$BC_{1}=b=A_{1}C_{1}$,则点$A_{1}(2+b$,$b)$,求出点$A_{1}$的横坐标为$sqrt{2}+1$,同理得出点$A_{2}$的横坐标为$sqrt{3}+sqrt{2}$;点$A_{3}$的横坐标为$sqrt{4}+sqrt{3}$;点$A_{4}$的横坐标为$sqrt{5}+sqrt{4}$;点的横坐标为$sqrt{6}+sqrt{5}$;根据规律可得答案.【解析】解:如图,过点$A、A_{1}、A_{2}、A_{3}cdots$分别作$ACperpx$轴,$A_{1}C_{1}perpx$轴,$A_{2}C_{2}perpx$轴,$A_{3}C_{3}perpx$轴$cdots$,垂足分别为$C、C_{1}、C_{2}、C_{3}cdotscdots$<ImageHere>$because$直线$OA$的关系式为$y=x,OAperpAB$,$thereforetriangleAOB$是等腰直角三角形,$thereforeOC=AC$,同理可得$triangleA_{1}BB_{1}、triangleA_{2}B_{1}B_{2}、triangleA_{3}B_{2}B_{3}cdotscdots$都是等腰直角三角形,设$OC=a=AC$,则点$A(a,a)$,点$A$在反比例函数的图象上,$thereforeatimesa=1$,解得:$a=1$(负值舍去),$therefore$点$A$的横坐标为1,设$A_{1}D=b$,则点$A_{1}(2+b,b)$,点$A_{1}$在反比例函数${y}=frac{1}{{x}}$的图象上,$therefore(2+b)timesb=1$,解得:$b=sqrt{2}-1$,$therefore$点$A_{1}$的横坐标为$2+sqrt{2}-1=sqrt{2}+1$;设$B_{1}C_{2}=c=A_{2}C_{2}$,则点${A}_{2}(2sqrt{2}+{c},c)$,点$A_{2}$在反比例函数${y}=frac{1}{{x}}$的图象上,$therefore(2sqrt{2}+c)timesc=1$,解得:$b=sqrt{3}-sqrt{2}$,$therefore$点$A_{2}$的横坐标为$sqrt{3}+sqrt{2};$同理可得:点$A_{3}$的横坐标为$sqrt{4}+sqrt{3}$;点$A_{4}$的横坐标为$sqrt{5}+sqrt{4};$点$A_{5}$的横坐标为$sqrt{6}+sqrt{5};$$cdotsldots$$therefore$点$A_{2023}$的横坐标为:$sqrt{2024}+sqrt{2023}$;故答案为:$sqrt{2024}+sqrt{2023}$.【点睛】本题考查反比例函数与一次函数图象的交点,掌握一次函数、反比例函数图象上点的坐标特征以及等腰直角三角形的性质是正确解答的关键.