·ÖÎö £¨1£©¸ù¾Ýº¯ÊýÖµÏàµÈµÄÁ½µã¹ØÓÚ¶Ô³ÆÖá¶Ô³Æ£¬¿ÉµÃA¡¢Bµã×ø±ê£¬¸ù¾Ý´ý¶¨ÏµÊý·¨£¬¿ÉµÃº¯Êý½âÎöʽ£»
£¨2£©¸ù¾ÝÏûÔª½â·½³Ì×飬¿ÉµÃ5x2+23x+9m-45=0£¬¸ù¾Ý¸ùÓëϵÊýµÄ¹ØÏµ£¬¿ÉµÃ£¨x1-x2£©2=£¨x1+x2£©2-4x1x2£¬¸ù¾Ý¹´¹É¶¨Àí£¬¿ÉµÃ¹ØÓÚmµÄ·½³Ì£¬¸ù¾Ý½â·½³Ì£¬¿ÉµÃ´ð°¸£»
£¨3£©¸ù¾Ý¹´¹É¶¨Àí£¬¿ÉµÃMN2=£¨n+2£©2+£¨5+5£©2£¬ME2=£¨n+5£©2+52£¬NE2=£¨n+3-n£©2+52=34£¬¸ù¾Ý¹´¹É¶¨ÀíµÄÄæ¶¨Àí£¬¿ÉµÃ¹ØÓÚnµÄ·½³Ì£¬¸ù¾Ý½â·½³Ì£¬¿ÉµÃnµÄÖµ£¬¿ÉµÃCµã×ø±ê£®
½â´ð ½â£º£¨1£©Å×ÎïÏßy=a£¨x+2£©2-5£¬µÃ
¶Ô³ÆÖáΪx=-2£®
ÓÉÅ×ÎïÏßy=a£¨x+2£©2-5ÓëxÖáÏཻÓÚA¡¢BÁ½µã£¬ÇÒAB=6£¬µÃ
-2+3=1£¬¼´B£¨1£¬0£©£¬-2-3=-5£¬¼´A£¨-5£¬0£©£¬
½«Aµã×ø±ê´úÈ뺯Êý½âÎöʽ£¬µÃ
9a-5=0£¬½âµÃm=$\frac{5}{9}$£¬
Å×ÎïÏߵĽâÎöʽy=$\frac{5}{9}$£¨x+2£©2-5£¬¶¥µãD£¨-2£¬-5£©£»
£¨2£©Èçͼ1
£¬
ÉèMNµÄ½âÎöʽΪy=-$\frac{1}{3}$x-m£¬M£¨x1£¬y1£©£¬N£¨x2£¬y2£©£®
ÁªÁ¢MNÓëÅ×ÎïÏߣ¬µÃ
$\left\{\begin{array}{l}{y=\frac{5}{9}£¨x+2£©^{2}-5}\\{y=-\frac{1}{3}x-m}\end{array}\right.$£¬
»¯¼ò£¬µÃ
5x2+23x+9m-45=0£®
x1+x2=-$\frac{23}{5}$£¬x1x2=$\frac{9m-45}{5}$£®
£¨x1-x2£©2=£¨x1+x2£©2-4x1x2=£¨-$\frac{23}{5}$£©2-4¡Á$\frac{9m-45}{5}$£®
£¨y1-y2£©2=£¨kx1-kx2£©2=k2£¨x1+x2£©2=£¨-$\frac{1}{3}$£©2[£¨-$\frac{23}{5}$£©2-4¡Á$\frac{9m-45}{5}$]
ÓÉMN=$\sqrt{10}$$\sqrt{£¨{x}_{1}-{x}_{2}£©^{2}+£¨{y}_{1}-{y}_{2}£©^{2}}$£¬µÃ
£¨-$\frac{23}{5}$£©2-4¡Á$\frac{9m-45}{5}$+£¨-$\frac{1}{3}$£©2[£¨-$\frac{23}{5}$£©2-4¡Á$\frac{9m-45}{5}$]=10£¬
»¯¼ò£¬µÃ180m=1204£¬
½âµÃm=$\frac{301}{45}$£»
£¨3£©ÓÉÐýתµÄÐÔÖÊ£¬µÃ
C£¨n£¬5£©£¬F£¨n+3£¬0£©£¬P£¨n-3£¬0£©£®
F¡¢A¹ØÓÚPµã¶Ô³Æ£¬µÃ
µã×ø±ê£¨$\frac{n-2}{2}$£¬0£©£®
DC2=£¨n+2£©2+£¨5+5£©2£¬DF2=£¨n+5£©2+52£¬CF2=£¨n+3-n£©2+52=34£»
¢Ùµ±CD2+DF2=CF2ʱ£¬£¨n+2£©2+£¨5+5£©2+£¨n+5£©2+52=34£¬
»¯¼ò£¬µÃn2+7n+60=0£¬¡÷=72-4¡Á1¡Á60=-191£¼0£¬·½³ÌÎ޽⣻
¢ÚÈçͼ2
£¬
µ±CD2+CF2=DF2ʱ£¬£¨n+2£©2+£¨5+5£©2+34=£¨n+5£©2+52£¬
»¯¼ò£¬µÃ6n=88£¬½âµÃn=$\frac{44}{3}$£¬
$\frac{n-2}{2}$=$\frac{\frac{44}{3}-2}{2}$=$\frac{19}{3}$£¬
´ËʱCµã×ø±êΪ£¨$\frac{19}{3}$£¬0£©£»
¢ÛÈçͼ3
£¬
µ±CF2+DF2=CD2ʱ£¬£¨n+5£©2+52+34=£¨n+2£©2+£¨5+5£©2£¬
»¯¼ò£¬µÃ6n=20£¬
½âµÃn=$\frac{10}{3}$£¬
$\frac{n-2}{2}$=$\frac{\frac{10}{3}-2}{2}$=$\frac{2}{3}$£¬
´ËʱCµã×ø±êΪ£¨$\frac{2}{3}$£¬0£©£®
×ÛÉÏËùÊö£ºÈôÒÔµãM¡¢N¡¢EΪ¶¥µãµÄÈý½ÇÐÎÊÇÖ±½ÇÈý½ÇÐÎʱ£¬µãCµÄ×ø±ê£¨$\frac{19}{3}$£¬0£©£¬£¨$\frac{2}{3}$£¬0£©£®
µãÆÀ ±¾Ì⿼²éÁ˶þ´Îº¯Êý×ÛºÏÌ⣬ÀûÓú¯ÊýÖµÏàµÈµÄÁ½µã¹ØÓÚ¶Ô³ÆÖá¶Ô³ÆµÃ³öA¡¢Bµã×ø±êÊǽâÌâ¹Ø¼ü£¬ÓÖÀûÓÃÁË´ý¶¨ÏµÊý·¨Çóº¯Êý½âÎöʽ£»ÀûÓôúÈëÏûÔª·¨µÃ³ö5x2+23x+9m-45=0ÊǽâÌâ¹Ø¼ü£¬ÓÖÀûÓÃÁ˹´¹É¶¨ÀíµÃ³ö¹ØÓÚmµÄ·½³Ì£»ÀûÓÃÁËÐýתµÄÐÔÖÊ£¬ÀûÓù´¹É¶¨ÀíµÃ³ö¹ØÓÚnµÄ·½³ÌÊǽâÌâ¹Ø¼ü£¬Òª·ÖÀàÌÖÂÛ£¬ÒÔ·ÀÒÅ©
| Äê¼¶ | ¸ßÖÐ¿Î³Ì | Äê¼¶ | ³õÖÐ¿Î³Ì |
| ¸ßÒ» | ¸ßÒ»Ãâ·Ñ¿Î³ÌÍÆ¼ö£¡ | ³õÒ» | ³õÒ»Ãâ·Ñ¿Î³ÌÍÆ¼ö£¡ |
| ¸ß¶þ | ¸ß¶þÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ | ³õ¶þ | ³õ¶þÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ |
| ¸ßÈý | ¸ßÈýÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ | ³õÈý | ³õÈýÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ |
¿ÆÄ¿£º³õÖÐÊýѧ À´Ô´£º ÌâÐÍ£º½â´ðÌâ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º³õÖÐÊýѧ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ
| A£® | 8$\sqrt{3}$ | B£® | 6 | C£® | 4$\sqrt{3}$ | D£® | 8 |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º³õÖÐÊýѧ À´Ô´£º ÌâÐÍ£º½â´ðÌâ
²é¿´´ð°¸ºÍ½âÎö>>
¿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
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º³õÖÐÊýѧ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ
| A£® | ÏàÀë | B£® | Ïཻ | ||
| C£® | ÏàÇÐ | D£® | ÒÔÉÏÈýÖÖÇé¿ö¾ùÓпÉÄÜ |
²é¿´´ð°¸ºÍ½âÎö>>
¹ú¼ÊѧУÓÅÑ¡ - 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