有个数学题,谁会帮帮忙。。
1个回答
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解:延长BC至G,使DC∥AC∵AD∥BC∴四边形ADGC为平行四边形∴DG=AC∵AC⊥BD∴DC⊥BD∵等腰梯形ABCD∴AC=BD∴DG=BD∴△DBG为等腰直角三角形∴BG2=2BD2∴(BC
AD)2=2BD2∴BD=DG=62∵DF⊥BG∴DF=FG∴2DF2=(6*根号2)2∴DF=6∴FC=6-4=2∵AE⊥BC,DF⊥BC,AD∥BC∴ADFE为矩形∴AE=DF,AD=EF∵AB=CD,∠AEB=∠DFC∴△ABE≌△DCF∴BE=DF∴EF=BC-2FC=8-2FC=4∴AE
EF=6
4=10.故选B.
1、BD=2x,则AD=8-2x,由8-2x≥0,得,x≤4,∴0≤x≤4
∵DE∥BC
∴AE/AC=AD/AB
即y/6=(8-2x)/8
∴y=6-3x/2(0≤x≤4)
2、S=1/2×BD×AE
=1/2×2x×(6-3x/2)
=-3/2x
6x
=-3/2(x-4x
4)
6
=-3/2(x-2)
6
∴当x=2时,Smax=6
2、(1)证明:∵AD=DC,DF⊥AC
∴∠DAC=∠DCA,AF=FC
∴∠FDC
∠DCA=90°
又∵∠DAB=∠DAC
∠CAB=90°
∴∠CAB=∠FDC
∴Rt△CDF∽Rt△BAC
∴拿岁AB/DC=BC/CF
∴AB×CF=CB×DC
即AB×AF=CB×CD
由题意,设x1=(1/2)^m1,x2=(1/2)^m2,m1,m2∈N
x1x2=(1/2)^m1*(1/2)^m2=(1/2)^(m1
m2),m1
m2必然是自然数,则x1x2∈A,选B
x1/x2=(1/2)^m1/(1/2)^m2=(1/2)^(m1-m2),m1-m2可能是负数就不是唤郑自然数了,切不可选x1/x2∈A
x1
x2,x1-x2就更和敏颂加不对了
AD)2=2BD2∴BD=DG=62∵DF⊥BG∴DF=FG∴2DF2=(6*根号2)2∴DF=6∴FC=6-4=2∵AE⊥BC,DF⊥BC,AD∥BC∴ADFE为矩形∴AE=DF,AD=EF∵AB=CD,∠AEB=∠DFC∴△ABE≌△DCF∴BE=DF∴EF=BC-2FC=8-2FC=4∴AE
EF=6
4=10.故选B.
1、BD=2x,则AD=8-2x,由8-2x≥0,得,x≤4,∴0≤x≤4
∵DE∥BC
∴AE/AC=AD/AB
即y/6=(8-2x)/8
∴y=6-3x/2(0≤x≤4)
2、S=1/2×BD×AE
=1/2×2x×(6-3x/2)
=-3/2x
6x
=-3/2(x-4x
4)
6
=-3/2(x-2)
6
∴当x=2时,Smax=6
2、(1)证明:∵AD=DC,DF⊥AC
∴∠DAC=∠DCA,AF=FC
∴∠FDC
∠DCA=90°
又∵∠DAB=∠DAC
∠CAB=90°
∴∠CAB=∠FDC
∴Rt△CDF∽Rt△BAC
∴拿岁AB/DC=BC/CF
∴AB×CF=CB×DC
即AB×AF=CB×CD
由题意,设x1=(1/2)^m1,x2=(1/2)^m2,m1,m2∈N
x1x2=(1/2)^m1*(1/2)^m2=(1/2)^(m1
m2),m1
m2必然是自然数,则x1x2∈A,选B
x1/x2=(1/2)^m1/(1/2)^m2=(1/2)^(m1-m2),m1-m2可能是负数就不是唤郑自然数了,切不可选x1/x2∈A
x1
x2,x1-x2就更和敏颂加不对了
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