Question

3£®Ö±Ïßy=-x+mÓëË«ÇúÏßy=$\frac{k}{x}$½»ÓÚµãE£¨$\frac{1}{2}$£¬2£©£¬FÁ½µã£®
£¨1£©Çók£¬mµÄÖµ¼°µãFµÄ×ø±ê£»
£¨2£©½«Ö±Ïßy=-x+mÑØyÖáÏòÏÂƽÒÆn¸öµ¥Î»ºóÇ¡ºÃÓëË«ÇúÏßy=$\frac{k}{x}$Ö»ÓÐÒ»¸ö½»µã£¬ÇónµÄÖµ£»
£¨3£©ÒÑÖªº¯Êýy=$\frac{1}{|x|}$µÄͼÏóÔÚµÚÒ»ÏóÏÞµÄÒ»Ö§ÇúÏßÉÏÓÐÒ»µãA£¨a£¬c£©£¬µãB£¨b£¬c+1£©Ôڸú¯ÊýͼÏóµÄÁíÍâÒ»Ö§ÉÏ£¬Éè¹ØÓÚxµÄÒ»Ôª¶þ´Î·½³Ìax²+bx+c=0µÄÁ½¸ùΪx₁£¬x₂£¬Çóx₁+x₂µÄÈ¡Öµ·¶Î§£®

Answer 1

·ÖÎö £¨1£©ÀûÓôý¶¨ÏµÊý·¨¼´¿ÉÇó³ök£¬mµÄÖµ£¬Í¨¹ý½â·½³Ì×é¼´¿ÉµÃ³öµãFµÄ×ø±ê£»
£¨2£©½áºÏ£¨1£©ÖеýáÂ۵óöƽÒƺóµÄÖ±ÏߵĽâÎöʽ£¬½«Æä´úÈë·´±ÈÀýº¯Êý½âÎöʽÖУ¬ÕûÀíµÃ³ö¹ØÓÚxµÄ¶þ´Î·½³Ì£¬ÁîÆä¸ùµÄÅбðʽ¡÷=0£¬¼´¿ÉµÃ³ö¹ØÓÚmµÄÒ»Ôª¶þ´Î·½³Ì£¬½â·½³Ì¼´¿ÉµÃ³ö½áÂÛ£»
£¨3£©¸ù¾ÝµãA£¨a£¬c£©ÔÚµÚÒ»ÏóÏÞµÄÒ»Ö§ÇúÏßÉÏ£¬µÃ³öa£¾0£¬c£¾0£¬ÔÙµãB£¨b£¬c+1£©Ôڸú¯ÊýͼÏóµÄÁíÍâÒ»Ö§ÉÏ£¬µÃ³öb£¼0£¬c+1£¾0£¬ÔÙ¸ù¾Ýx₁+x₂=-$\frac{b}{a}$£¬¼´¿ÉµÃ³öx₁+x₂µÄÈ¡Öµ·¶Î§£®

½â´ð ½â£º£¨1£©¡ßÖ±Ïßy=-x+mÓëË«ÇúÏßy=$\frac{k}{x}$½»ÓÚµãE£¨$\frac{1}{2}$£¬2£©£¬
¡à$\left\{\begin{array}{l}{2=-\frac{1}{2}+m}\\{2=\frac{k}{\frac{1}{2}}}\end{array}\right.$£¬½âµÃ$\left\{\begin{array}{l}{m=\frac{5}{2}}\\{k=1}\end{array}\right.$£¬
½â·½³Ì×é$\left\{\begin{array}{l}{y=-x+\frac{5}{2}}\\{y=\frac{1}{x}}\end{array}\right.$£¬
¿ÉµÃ$\left\{\begin{array}{l}{x=2}\\{y=\frac{1}{2}}\end{array}\right.$»ò$\left\{\begin{array}{l}{x=\frac{1}{2}}\\{y=2}\end{array}\right.$£¬
¡àµãFµÄ×ø±êΪ£¨2£¬$\frac{1}{2}$£©£»

£¨2£©½«Ö±Ïßy=-x+$\frac{5}{2}$ÑØyÖáÏòÏÂƽÒÆn¸öµ¥Î»ºó£¬¿ÉµÃy=-x+$\frac{5}{2}$-n£¬
ÓÉ-x+$\frac{5}{2}$-n=$\frac{1}{x}$£¬¿ÉµÃx²+£¨n-$\frac{5}{2}$£©x+1=0£¬
¡ßƽÒƺóµÄÖÐÏßÇ¡ºÃÓëË«ÇúÏßy=$\frac{k}{x}$Ö»ÓÐÒ»¸ö½»µã£¬
¡à¡÷=£¨n-$\frac{5}{2}$£©²-4=0£¬
½âµÃn=$\frac{1}{2}$»ò$\frac{9}{2}$£»

£¨3£©¡ßµãA£¨a£¬c£©ÔÚµÚÒ»ÏóÏÞµÄÒ»Ö§ÇúÏßÉÏ£¬
¡àa£¾0£¬c£¾0£¬ÇÒac=1£¬¼´a=$\frac{1}{c}$£¬
¡ßµãB£¨b£¬c+1£©Ôڸú¯ÊýͼÏóµÄÁíÍâÒ»Ö§ÉÏ£¬¼´µÚ¶þÏóÏÞ£¬
¡àb£¼0£¬c+1£¾0£¬ÇÒb£¨c+1£©=-1£¬¼´b=-$\frac{1}{c+1}$£¬
¡àx₁+x₂=-$\frac{b}{a}$=$\frac{c}{c+1}$£¬
¡à0£¼x₁+x₂£¼1£®

µãÆÀ ±¾Ì⿼²éÁË·´±ÈÀýº¯ÊýÓëÒ»´Îº¯ÊýµÄ½»µãÎÊÌâ¡¢´ý¶¨ÏµÊý·¨Çóº¯Êý½âÎöʽ¡¢¸ùµÄÅбðʽÒÔ¼°½âÒ»Ôª¶þ´Î·½³Ì£¬½âÌâµÄ¹Ø¼üÊÇ£ºÀûÓôý¶¨ÏµÊý·¨Çóº¯Êý½âÎöʽ£»ÀûÓøùµÄÅбðʽµÃ³ö¹ØÓÚnµÄÒ»Ôª¶þ´Î·½³Ì£®