,由b 2+c 2=25得,0<c 2<25,∴0<c 2<16Р两式相加,得a 2+b 2+2c 2=41,a 2+b 2=41-2c 2Р由0<c 2<16得9<41-2c 2<41,即9<a 2+b 2<41Р11.60°<∠A<90°Р解:∵BD=AB=AC,∴∠ADB=∠A,∠C=(180°-∠A)Р∵∠ADB>∠C,∴∠A>(180°-∠A),∴∠A>60°Р由∠A+∠ADB<180°,得2∠A<180°,A<90°Р故60°<∠A<90°РxРyРOР12.-1Р(x≥0)Р(x≤0)Р解:y=2x 2+4|x|-1=2(|x|+1)2-3=Р其图象如图,由图象可知,当x=0时,y最小为-1Р13.<Р解:由题意得:y1=ax 12+2ax1+4,y2=ax 22+2ax2+4Рy1-y2=a(x 12-x 22)+2a(x 1-x 2)=a(x 1-x 2)(x 1+x 2+2)=a(x 1-x 2)(3-a)Р∵x1<x2,0< a <3,∴y1-y2<0,∴y1<y2Р14.Р解:过C作CE⊥AB于E,过D作DF⊥AB于F,DG⊥AC于GРAРDРBРCРEРFРGР∵S△ABC =AB·CE=AB·AC·sin60°РS△ABC =S△ABD+S△ADC =AB·DF+AC·DG=AB·AD·sin30°+AC·AD·sin30°Р∴AB·AC·sin60°=AB·AD·sin30°+AC·AD·sin30°Р解得AD=Р15.y=-x 2+x-,<x<10Р解:AB2=AC 2+BC 2=6 2+8 2=100,AB=10Р由△ADE∽△ABC得DE=x,AE=x,CE=6-xР由△BFD∽△ABC得BF=-x,CF=8-(-x)=x-Рy=(CF+DE)·CE=(x-+x)(6-x)=-x 2+x-Р当点F与点C重合时,由△ACD∽△ABC得AD=
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