·ÖÎö £¨1£©¹ýµãC×÷CD¡ÍxÖáÓÚµãD£¬ÀûÓù´¹É¶¨Àí¿ÉµÃCO=5£¬È»ºó¸ù¾ÝÕÛµþµÄÐÔÖÊ¿ÉÖªAO=5£¬ÔÙÀûÓôý¶¨ÏµÊý·¨¿ÉÇó³öÅ×ÎïÏߵĽâÎöʽ£»
£¨2£©Ê×ÏÈÈ·¶¨OCµÄ½âÎöʽ£¬½ø¶øÈ·¶¨Æ½ÐÐÓÚOCµÄÖ±Ïß½âÎöʽ£¬È»ºóÁªÁ¢½âÎöʽ£¬µÃ³ö¹ØÓÚmµÄ·½³Ì£¬ÀûÓÃÅбðʽΪ0Çó³ömµÄÖµ£¬ÔÙÀûÓáÏOCDµÄÕýÏÒÇó½â£¬¿ÉµÃ½áÂÛ£»
£¨3£©·Öµ±PÔÚxÖáÉÏ·½£¬µ±PÔÚxÖáÏ·½Á½ÖÖÇé¿öÌÖÂۿɵõãPµÄ×ø±ê£®
½â´ð ½â£º£¨1£©ÈçͼËùʾ£º¹ýµãC×÷CD¡ÍxÖáÓÚµãD£¬
¡ßC£¨3£¬4£©£¬
¡àDO=3£¬DC=4£¬
¡àCO=5£¬
¡ß½«Rt¡÷OABÑØOBÕÛµþºó£¬µãAÂäÔÚµÚÒ»ÏóÏÞÄڵĵãC´¦£¬
¡àAO=5£¬
ÉèÅ×ÎïÏß½âÎöʽΪ£ºy=ax2+bx+c£¬
Ôò$\left\{\begin{array}{l}{c=0}\\{9a+3b+c=4}\\{25a+5b+c=0}\end{array}\right.$£¬
½âµÃ£º$\left\{\begin{array}{l}{a=-\frac{2}{3}}\\{b=\frac{10}{3}}\\{c=0}\end{array}\right.$£¬
¹ÊÅ×ÎïÏß½âÎöʽΪ£ºy=-$\frac{2}{3}$x2+$\frac{10}{3}$x£»
£¨2£©¡ßC£¨3£¬4£©£¬
¡àÖ±ÏßOCµÄ½âÎöʽΪy=$\frac{4}{3}$x£¬
ÉèµãMµ½OCµÄ×î´ó¾àÀëʱ£¬Æ½ÐÐÓÚOCµÄÖ±Ïß½âÎöʽΪy=$\frac{4}{3}$x+m£¬
ÁªÁ¢$\left\{\begin{array}{l}{y=\frac{4}{3}x+m}\\{y=-\frac{2}{3}{x}^{2}+\frac{10}{3}x}\end{array}\right.$£¬
Ïûµôδ֪Êýy²¢ÕûÀíµÃ2x2-6x+3m=0£¬
¡÷=£¨-6£©2-24m=0£¬
½âµÃ£ºm=$\frac{3}{2}$£®
ÉèµãMµ½OCµÄ×î´ó¾àÀëΪx£¬
¡ß¡ÏNOM=¡ÏOCD£¬
¡àsin¡ÏOCD=sin¡ÏNOM=$\frac{3}{5}$£¬
Ôò$\frac{x}{\frac{3}{2}}$=$\frac{3}{5}$£¬
½âµÃx=$\frac{9}{10}$£»
£¨3£©ACµÄÖеã×ø±êΪ£¨4£¬2£©£¬
ÔòÖ±ÏßOB½âÎöʽΪy=$\frac{1}{2}$x£¬
ÁªÁ¢$\left\{\begin{array}{l}{y=\frac{1}{2}x}\\{y=-\frac{2}{3}{x}^{2}+\frac{10}{3}x}\end{array}\right.$£¬
½âµÃ$\left\{\begin{array}{l}{{x}_{1}=0}\\{{y}_{1}=0}\end{array}\right.$£¬$\left\{\begin{array}{l}{{x}_{2}=\frac{17}{4}}\\{{y}_{2}=\frac{17}{8}}\end{array}\right.$£¬
µ±PÔÚxÖáÉÏ·½£¬µãPµÄ×ø±êΪ£¨$\frac{3}{4}$£¬$\frac{17}{8}$£©£»
µ±PÔÚxÖáÉÏ·½£¬
BµÄ×ø±êΪ£¨5£¬$\frac{5}{2}$£©£¬
APµÄ½âÎöʽΪy=$\frac{1}{2}$x-$\frac{5}{2}$£¬
ÁªÁ¢$\left\{\begin{array}{l}{y=\frac{1}{2}x-\frac{5}{2}}\\{y=-\frac{2}{3}{x}^{2}+\frac{10}{3}x}\end{array}\right.$£¬
½âµÃ$\left\{\begin{array}{l}{{x}_{1}=5}\\{{y}_{1}=0}\end{array}\right.$£¨ÉáÈ¥£©£¬$\left\{\begin{array}{l}{{x}_{2}=-\frac{3}{4}}\\{{y}_{2}=-\frac{23}{8}}\end{array}\right.$£¬
µãPµÄ×ø±êΪ£¨-$\frac{3}{4}$£¬-$\frac{23}{8}$£©£®
×ÛÉÏËùÊö£¬´æÔÚÒ»µãP£¨$\frac{3}{4}$£¬$\frac{17}{8}$£©»ò£¨-$\frac{3}{4}$£¬-$\frac{23}{8}$£©£¬Ê¹¡ÏOAP=¡ÏBOA£®
µãÆÀ ±¾Ì⿼²éÁ˶þ´Îº¯ÊýµÄ×ÛºÏÓ¦Óã¬Éæ¼°ÁËͼÐεķÕ۱任£¬ÀûÓôý¶¨ÏµÊý·¨ÇóÒ»´Îº¯Êý¡¢¶þ´Îº¯ÊýµÄ½âÎöʽ£¬Ö±ÏßÓëÅ×ÎïÏߵĽ»µãµÄÇ󷨵È֪ʶ£¬¾ßÓÐÒ»¶¨µÄ×ÛºÏÐÔÓëÄѶȣ¬½âÌâʱҪעÒâÊýÐνáºÏ˼ÏëÓë·½³Ì˼ÏëµÄÓ¦Óã®
Ä꼶 | ¸ßÖÐ¿Î³Ì | Ä꼶 | ³õÖÐ¿Î³Ì |
¸ßÒ» | ¸ßÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ | ³õÒ» | ³õÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ |
¸ß¶þ | ¸ß¶þÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õ¶þ | ³õ¶þÃâ·Ñ¿Î³ÌÍƼö£¡ |
¸ßÈý | ¸ßÈýÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õÈý | ³õÈýÃâ·Ñ¿Î³ÌÍƼö£¡ |
¿ÆÄ¿£º³õÖÐÊýѧ À´Ô´£º ÌâÐÍ£º½â´ðÌâ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º³õÖÐÊýѧ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º³õÖÐÊýѧ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º³õÖÐÊýѧ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
ÐòºÅ | ¢Ù | ¢Ú | ¢Û | ¢Ü |
Öܳ¤ | 6 | 10 | 16 | 26 |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º³õÖÐÊýѧ À´Ô´£º ÌâÐÍ£º½â´ðÌâ
²Ù×÷´ÎÊýN | 1 | 2 | 3 | 4 | 5 | ¡ | n |
Õý·½ÐθöÊý | 4 | 7 | 10 | an |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º³õÖÐÊýѧ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º³õÖÐÊýѧ À´Ô´£º ÌâÐÍ£º½â´ðÌâ
²é¿´´ð°¸ºÍ½âÎö>>
°Ù¶ÈÖÂÐÅ - Á·Ï°²áÁбí - ÊÔÌâÁбí
ºþ±±Ê¡»¥ÁªÍøÎ¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨Æ½Ì¨ | ÍøÉÏÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | µçÐÅթƾٱ¨×¨Çø | ÉæÀúÊ·ÐéÎÞÖ÷ÒåÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | ÉæÆóÇÖȨ¾Ù±¨×¨Çø
Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com