8£®ÔÚÆ½ÃæÖ±½Ç×ø±êϵxOyÖУ¬Ô²CµÄ·½³ÌΪ£¨x-4£©2+y2=1£¬ÇÒÔ²CÓëxÖá½»ÓÚM£¬NÁ½µã£¬ÉèÖ±ÏßlµÄ·½³ÌΪy=kx£¨k£¾0£©
£¨1£©µ±Ö±ÏßlÓëÔ²CÏàÇÐʱ£¬ÇóÖ±ÏßlµÄ·½³Ì£»
£¨2£©ÒÑÖªÖ±ÏßlÓëÔ²CÏཻÓÚA£¬BÁ½µã
£¨i£©ÈôAB¡Ü$\frac{2\sqrt{17}}{17}$£¬ÇóʵÊýkµÄȡֵ·¶Î§£»
£¨ii£©Ö±ÏßAMÓëÖ±ÏßBNÏཻÓÚµãP£¬Ö±ÏßAM£¬Ö±ÏßBN£¬Ö±ÏßOPµÄбÂÊ·Ö±ðΪk1£¬k2£¬k3£¬ÊÇ·ñ´æÔÚ³£Êýa£¬Ê¹µÃk1+k2=ak3ºã³ÉÁ¢£¿Èô´æÔÚ£¬Çó³öaµÄÖµ£»Èô²»´æÔÚ£¬ËµÃ÷ÀíÓÉ£®

·ÖÎö £¨1£©ÓÉÌâÒâk£¾0£¬Ô²ÐÄCµ½Ö±ÏßlµÄ¾àÀëd=$\frac{4k}{\sqrt{1+{k}^{2}}}$£¬ÓÉÖ±ÏßlÓëÔ²CÏàÇеÃk=$\frac{\sqrt{15}}{15}$£¬ÓÉ´ËÄÜÇó³öÖ±Ïßl£®
£¨2£©£¨i£©ÓÉÌâÒâµÃ0£¼AB=2$\sqrt{1-{d}^{2}}$¡Ü$\frac{2\sqrt{17}}{17}$£¬´Ó¶ød=$\frac{4k}{\sqrt{1+{k}^{2}}}$£¬ÓÉ´ËÄÜÇó³öʵÊýkµÄȡֵ·¶Î§£®
£¨ii£©lAM£ºy=k1£¨x-3£©£¬ÓëÔ²C£º£¨x-4£©2+y2=1ÁªÁ¢£¬µÃ$£¨x-3£©[£¨1+{{k}_{1}}^{2}£©x-£¨3{{k}_{1}}^{2}+5£©]=0$£¬ÓÉΤ´ï¶¨ÀíÇó³öA£¬BµÄ×ø±ê£¬´Ó¶øµÃµ½£¨1+k1k2£©£¨3k1+5k2£©=0£¬ÓÉ´ËÄÜÖ¤Ã÷´æÔÚ³£Êýa=2£¬Ê¹µÃk1+k2=2k3ºã³ÉÁ¢£®

½â´ð ½â£º£¨1£©ÓÉÌâÒâk£¾0£¬
¡àÔ²ÐÄCµ½Ö±ÏßlµÄ¾àÀëd=$\frac{4k}{\sqrt{1+{k}^{2}}}$£¬
¡ßÖ±ÏßlÓëÔ²CÏàÇУ¬
¡àd=$\frac{4k}{\sqrt{1+{k}^{2}}}$=1£¬½âµÃk=$\frac{\sqrt{15}}{15}$£¬
¡àÖ±Ïßl£ºy=$\frac{\sqrt{15}}{15}$x£®
£¨2£©£¨i£©ÓÉÌâÒâµÃ0£¼AB=2$\sqrt{1-{d}^{2}}$¡Ü$\frac{2\sqrt{17}}{17}$£¬
½âµÃ$\frac{4\sqrt{17}}{17}¡Üd£¼1$£¬
ÓÉ£¨1£©Öªd=$\frac{4k}{\sqrt{1+{k}^{2}}}$£¬
¡à$\frac{4\sqrt{17}}{17}¡Ü\frac{4k}{\sqrt{1+{k}^{2}}}£¼1$£¬
¡à$\frac{1}{4}¡Ük£¼\frac{\sqrt{15}}{15}$£®
£¨ii£©lAM£ºy=k1£¨x-3£©£¬
ÓëÔ²C£º£¨x-4£©2+y2=1ÁªÁ¢£¬
µÃ$£¨x-3£©[£¨1+{{k}_{1}}^{2}£©x-£¨3{{k}_{1}}^{2}+5£©]=0$£¬
¡à${x}_{M}=3£¬{x}_{A}=\frac{3{{k}_{1}}^{2}+5}{1+{{k}_{1}}^{2}}$£¬
¡àA£¨$\frac{3{{k}_{1}}^{2}+5}{1+{{k}_{1}}^{2}}$£¬$\frac{2{k}_{1}}{1+{{k}_{1}}^{2}}$£©£¬
ͬÀí£¬µÃB£¨$\frac{5{{k}_{2}}^{2}+3}{1+{{k}_{2}}^{2}}$£¬$\frac{-2{k}_{2}}{1+{{k}_{2}}^{2}}$£©£¬
¡ßkOA=kOB£¬
¡à$\frac{\frac{2{k}_{1}}{1+{{k}_{1}}^{2}}}{\frac{3{{k}_{1}}^{2}+5}{1+{{k}_{1}}^{2}}}$=$\frac{\frac{-2{k}_{2}}{1+{{k}_{2}}^{2}}}{\frac{5{{k}_{2}}^{2}+3}{1+{{k}_{2}}^{2}}}$£¬¼´£¨1+k1k2£©£¨3k1+5k2£©=0£¬
¡ßk1k2¡Ù-1£¬
¡à${k}_{2}=-\frac{3}{5}{k}_{1}$£¬
ÉèP£¨x0£¬y0£©£¬
Ôò$\left\{\begin{array}{l}{{y}_{0}={k}_{1}£¨{x}_{0}-3£©}\\{{y}_{0}={k}_{2}£¨{x}_{0}-5£©}\end{array}\right.$£¬
¡à$\left\{\begin{array}{l}{{x}_{0}=\frac{3{k}_{1}-5{k}_{2}}{{k}_{1}-{k}_{2}}}\\{{y}_{0}=\frac{-2{k}_{1}{k}_{2}}{{k}_{1}-{k}_{2}}}\end{array}\right.$£®
¡àP£¨$\frac{3{k}_{1}-5{k}_{2}}{{k}_{1}-{k}_{2}}$£¬$\frac{-2{k}_{1}{k}_{2}}{{k}_{1}-{k}_{2}}$£©£¬¼´P£¨$\frac{15}{4}$£¬$\frac{3{k}_{1}}{4}$£©£¬
¡à${k}_{3}=\frac{\frac{3{k}_{1}}{4}}{\frac{15}{4}}$=$\frac{{k}_{1}}{5}$£¬
¡à${k}_{1}+{k}_{2}=\frac{2}{5}{k}_{1}=2{k}_{3}$£¬
¡à´æÔÚ³£Êýa=2£¬Ê¹µÃk1+k2=2k3ºã³ÉÁ¢£®

µãÆÀ ±¾Ì⿼²éÖ±Ïß·½³ÌµÄÇ󷨣¬¿¼²éÖ±ÏßµÄбÂʵÄȡֵ·¶Î§µÄÇ󷨣¬¿¼²éÊÇ·ñ´æÔÚʹµÃµÈʽºã³ÉÁ¢µÄ³£ÊýµÄÅжÏÓëÖ¤Ã÷£¬ÊÇÖеµÌ⣬½âÌâʱҪÈÏÕæÉóÌ⣬עÒâÔ²µÄÐÔÖʵĺÏÀíÔËÓã®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖÐÊýѧ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

18£®µ×Ãæ°ë¾¶Îª$\sqrt{3}$£¬Ä¸Ïß³¤Îª2µÄÔ²×¶µÄÍâ½ÓÇòOµÄ±íÃæ»ýΪ£¨¡¡¡¡£©
A£®6¦ÐB£®12¦ÐC£®8¦ÐD£®16¦Ð

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖÐÊýѧ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

19£®ÏÂÁÐÄĸöº¯ÊýÊÇÆæº¯Êý£¨¡¡¡¡£©
A£®f£¨x£©=3x3+2x2+1B£®f£¨x£©=${x^{-\frac{1}{2}}}$C£®f£¨x£©=3xD£®f£¨x£©=$\frac{{\sqrt{4-{x^2}}}}{{|{x+3}|-3}}$

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖÐÊýѧ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

16£®Èçͼ³ÌÐòÔËÐкóµÄ½á¹ûÊÇ£¨¡¡¡¡£©
A£®A+2B£®2013C£®2014D£®2015

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖÐÊýѧ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

3£®ÔÚ¡÷ABCÖУ¬tanAÊÇÒÔ-4ΪµÚÈýÏ4ΪµÚÆßÏîµÄµÈ²îÊýÁеĹ«²î£¬tanBÊÇÒÔ2Ϊ¹«²î£¬9ΪµÚÎåÏîµÄµÈ²îÊýÁеĵڶþÏÔòÕâ¸öÈý½ÇÐÎÊÇ£¨¡¡¡¡£©
A£®Èñ½ÇÈý½ÇÐÎB£®¶Û½ÇÈý½ÇÐÎ
C£®µÈÑüÖ±½ÇÈý½ÇÐÎD£®µÈÑü»òÖ±½ÇÈý½ÇÐÎ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖÐÊýѧ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

13£®Éú²ú¹ý³ÌÓÐ4µÀ¹¤Ðò£¬Ã¿µÀ¹¤ÐòÐèÒª°²ÅÅÒ»ÈËÕÕ¿´£¬ÏִӼס¢ÒÒ¡¢±ûµÈ6Ãû¹¤ÈËÖа²ÅÅ4ÈË·Ö±ðÕÕ¿´Ò»µÀ¹¤Ðò£¬µÚÒ»µÀ¹¤Ðò°²ÅÅÒÒ×ö£¬µÚËĵÀ¹¤ÐòÖ»ÄܴӼס¢±ûÁ½ÈËÖа²ÅÅ1ÈË£¬Ôò²»Í¬µÄ°²ÅÅ·½°¸ÓжàÉÙÖÖ£¿

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖÐÊýѧ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

20£®8ÈËÅÅÁ½ÅÅ£¬ÅźóÁ½ÅŸ÷4ÈË£¬×é³É2¡Á4·½Õó£®
£¨1£©¼×¡¢ÒÒ²»Í¬ÅÅÓжàÉÙÅÅ·¨£¿
£¨2£©¼×¡¢ÒÒͬÅÅÓжàÉÙÅÅ·¨£¿
£¨3£©¼×¡¢ÒÒͬÅÅÏàÁÚ»òǰºóÏàÁÚÓжàÉÙÅÅ·¨£¿
£¨4£©¼×¡¢ÒÒ²»ÔÚÁ½¶ËÓжàÉÙÅÅ·¨£¿
£¨5£©ÈÎÒâÅÅÓжàÉÙÅÅ·¨£¿

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖÐÊýѧ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

17£®ÒÑÖªÖ±Ïßl£ºax-y+2=0ÓëÔ²M£ºx2+y2-4y+3=0µÄ½»µãΪA¡¢B£¬µãCÊÇÔ²MÉÏÒ»¶¯µã£¬ÉèµãP£¨0£¬-1£©£¬Ôò|$\overrightarrow{PA}$+$\overrightarrow{PB}$+$\overrightarrow{PC}$|µÄ×î´óֵΪ10£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖÐÊýѧ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

18£®´ÓijѧУ¸ßÈýÄê¼¶¹²800ÃûÄÐÉúÖÐËæ»ú³éÈ¡50È˲âÁ¿Éí¸ß£®Êý¾Ý±íÃ÷£¬±»²âѧÉúÉí¸ßÈ«²¿½éÓÚ155cmµ½195cmÖ®¼ä£¬½«²âÁ¿½á¹û°´ÈçÏ·½Ê½·Ö³É°Ë×飺µÚÒ»×é[155£¬160£©£»µÚ¶þ×é[160£¬165£©£»¡­£»µÚ°Ë×é[190£¬195]£®ÈçͼÊǰ´ÉÏÊö·Ö×é·½·¨µÃµ½µÄƵÂÊ·Ö²¼Ö±·½Í¼µÄÒ»²¿·Ö£®ÒÑÖªµÚÒ»×éÓëµÚ°Ë×éÈËÊýÏàͬ£¬µÚÁù×é±ÈµÚÆß×éÉÙ1ÈË£®
£¨1£©¹À¼ÆÕâËùѧУ¸ßÈýÄê¼¶È«ÌåÄÐÉúÉí¸ßÔÚ180cmÒÔÉÏ£¨º¬180cm£©µÄÈËÊý£»
£¨2£©Èô´ÓÉí¸ßÊôÓÚµÚÁù×éºÍµÚ°Ë×éµÄËùÓÐÄÐÉúÖÐËæ»ú³éÈ¡Á½ÈË£¬¼ÇËûÃǵÄÉí¸ß·Ö±ðΪx£¬y£¬ÇóÂú×ã¡°|x-y|¡Ü5¡±µÄʼþµÄ¸ÅÂÊ£®

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸