【導(dǎo)讀】17.解:(Ⅰ)△ABD中,由正弦定理,在△ACD中,由余弦定理:,整理得CD2+6CD-40=0,解得CD=-10(舍去),CD=4,………………∴S△ABC=.……………………18.解:(Ⅰ)設(shè){an}的公差為d(d>0),由S3=15有3a1+=15,化簡得a1+d=5,①………………………又∵a1,a4,a13成等比數(shù)列,∴a42=a1a13,即2=a1,化簡得3d=2a1,②……………(Ⅱ)∵+11,即,又≥6,當(dāng)且僅當(dāng)n=3時,等號成立,(Ⅱ)由(Ⅰ)知,③當(dāng)>1時,當(dāng)且僅當(dāng)=1時,取得最大值4-1,由題知方程=0恰有三個實數(shù)根,∴在上單調(diào)遞增,在上單調(diào)遞減.………且當(dāng)時,;當(dāng)時,,22.解:(Ⅰ)將C的參數(shù)方程化為普通方程為(x-3)2+(y-4)2=25,(Ⅱ)把代入,得,