【正文】
? ??? 18.(1) 23)62s i n(2 2c os112s i n2 3)( ??????? ?xxxxf ???T (2) 23)32sin()( ??? ?xxg 20 ???x? 32323 ??? ????? x 2 33)(0332x m in ???????? xgx 時(shí),即當(dāng) ?? 21)(125232x ax ????? mxgx 時(shí),即當(dāng) ??? ( 1) CbaB c o s)2(c o sc ??? CBCABC c o ss i nc o ss i n2c o ss i n ??? 由正弦定理可知, CABC CACBBC c oss i n2)s i n( c oss i n2c oss i nc oss i n ?? ?? ???? CBA? CAA c o ssin2sin ?? 3021c os0s i n ?? ??????? CCCA ?? (2)由題可知 3,4 ??? Cc abS ABC 43?? ? Cabcba c o s2222 ???由余弦定理可知:? abba ??? 1622 ”時(shí)等號(hào)成立當(dāng)且僅當(dāng)“ baabababba ??????? 1621622 34最大值是ABCS ?? 此時(shí)三角形為等邊三角形 20(1) 。時(shí),當(dāng) 1 112 221 ? ??? ????? nn nnnnnaa aaSSan 1 1211 111? ????a aSan 時(shí),當(dāng) 21}{a n 為公比的等比數(shù)列為首項(xiàng),是以? 1nn 2?a ( 2) 12n2lo glo gb 12n222n ??? na )12 112 1(21)12(12 1bb 1 1nn ???????? nnnn )( 1212112n1513131121n???????? ????????nnnT )()()( ? 21.( 1) 043)21(1 22 ?????? xxx? 03)1(1 14222????? ?????? xkxk xxkxkx )即( 成立恒成立。