Question

15£®¶ÔÓÚÕýʵÊýa£¬b£»£¨$\frac{a+b}{2}$£©²-ab=$\frac{{a}^{2}+2ab+{b}^{2}}{4}$-$\frac{4ab}{4}$=$\frac{{a}^{2}-2ab+{b}^{2}}{4}$=£¨$\frac{a-b}{2}$£©²
¡ß£¨$\frac{a-b}{2}$£©²¡Ý0£»¡àab¡Ü£¨$\frac{a+b}{2}$£©²£»Èçͼ£ºÈç¹ûµãA£¨a£¬b£©ÊÇ·´±ÈÀýº¯Êýy=$\frac{3}{x}$£¨x£¾0£©Í¼ÏóÉÏÒ»µã£¬¹ýµãA×÷xÖáµÄ´¹Ïߣ»¹ýµãA×÷yÖáµÄ´¹Ïߣ¬´¹×ã·Ö±ðΪµãB£¬C£¬·ÂÕÕÉÏÃæ¸ø¶¨µÄÌõ¼þ
£¨1£©Çó³öËıßÐÎOBACµÄÖܳ¤×îСֵ
£¨2£©ËıßÐÎOBACµÄÖܳ¤×îСʱµãAµÄ×ø±ê£®

Answer 1

·ÖÎö £¨1£©ÓɵãAÔÚµÚÒ»ÏóÏÞ¼´¿ÉµÃ³öa£¾0¡¢b£¾0¡¢a+b£¾0¡¢ab=3£¬½áºÏab¡Ü£¨$\frac{a+b}{2}$£©²¼´¿ÉµÃ³öa+b¡Ý2$\sqrt{3}$£¬ÔÙ¸ù¾Ý¾ØÐεÄÖܳ¤¹«Ê½¼´¿ÉµÃ³ö½áÂÛ£»
£¨2£©Óɵ±a=bʱa-b=0£¬¼´¿ÉµÃ³öµ±a=bʱËıßÐÎOBACµÄÖܳ¤×îС£¬½áºÏab=3¼´¿ÉµÃ³öµãAµÄ×ø±ê£®

½â´ð ½â£º£¨1£©¡ßµãA£¨a£¬b£©ÊÇ·´±ÈÀýº¯Êýy=$\frac{3}{x}$£¨x£¾0£©Í¼ÏóÉÏÒ»µã£¬
¡àa£¾0£¬b£¾0£¬a+b£¾0£¬ab=3£®
¡ßab¡Ü£¨$\frac{a+b}{2}$£©²£¬
¡àa+b¡Ý2$\sqrt{3}$£¬
¡àËıßÐÎOBACµÄÖܳ¤2£¨a+b£©¡Ý4$\sqrt{3}$£¬¼´ËıßÐÎOBACµÄÖܳ¤×îСֵΪ4$\sqrt{3}$£®
£¨2£©¡ßµ±a=bʱ£¬a-b=0£¬
¡àµ±a=bʱ£¬ËıßÐÎOBACµÄÖܳ¤×îС£®
¡ßab=3£¬ÇÒµãA£¨a£¬b£©ÔÚµÚÒ»ÏóÏÞ£¬
¡àËıßÐÎOBACµÄÖܳ¤×îСʱµãAµÄ×ø±êΪ£¨$\sqrt{3}$£¬$\sqrt{3}$£©£®

µãÆÀ ±¾Ì⿼²éÁË·´±ÈÀýº¯ÊýͼÏóÉϵãµÄ×ø±êÌØÕ÷¡¢²»µÈʽµÄÐÔÖÊÒÔ¼°¾ØÐεÄÖܳ¤£¬½âÌâµÄ¹Ø¼üÊÇ£º£¨1£©¸ù¾Ý²»µÈʽµÄÐÔÖÊÕÒ³öa+b¡Ý2$\sqrt{3}$£»£¨2£©¸ù¾Ý²»µÈʽµÄÐÔÖÊÕÒ³öµ±a=bʱ£¬ËıßÐÎOBACµÄÖܳ¤×îС£®