作业帮八上数学题!各位大神求解!
来源：学生作业帮 编辑：作业帮 分类：数学作业 时间：2024/06/11 01:26:13
八上数学题!各位大神求解!
1,
同时出发,速度一样,所以AP=BQ
∠B=∠CAP
AB=CA
∴△ABQ≌△CAP
2,
∵△ABQ≌△CAP
∴∠BAQ=∠ACP
∠CAQ+∠ACP=∠CAQ+∠BAQ=60°
∴∠QMP=∠AMC=180°-(∠CAQ+∠BAQ)=180°-60°=120°
3,BP=CQ
∠CBP=∠ACQ
BC=AC
∴△BPC≌△CQA
∴∠BPC=∠CQA
∴∠QMC=180°-(∠MCQ+AQC)
=180°-(∠BCP+∠BPC)
=180°-∠ABC
=120°
再问： 那31题了！谢谢
再答： (1): m^2=2n+1……① 4n^2=m+1……② ①-② m^2-4n^2=2n-m (m+2n)(m-2n)=2n-m m≠2n ∴m+2n=-1 (2): (m+2n)^2=(-1)^2=1 m^2+4mn+4n^2=1 2n+1+4mn+m+1=1 m+2n+4mn+1=0 -1+4mn+1=0 4mn=0 mn=0 ∴4n^3-mn+2n^2 =n(4n^2-m+2n) =n(m+1-m+2n) =n(2n+1) =nm^2 =nm×m =0×m =0
同时出发,速度一样,所以AP=BQ
∠B=∠CAP
AB=CA
∴△ABQ≌△CAP
2,
∵△ABQ≌△CAP
∴∠BAQ=∠ACP
∠CAQ+∠ACP=∠CAQ+∠BAQ=60°
∴∠QMP=∠AMC=180°-(∠CAQ+∠BAQ)=180°-60°=120°
3,BP=CQ
∠CBP=∠ACQ
BC=AC
∴△BPC≌△CQA
∴∠BPC=∠CQA
∴∠QMC=180°-(∠MCQ+AQC)
=180°-(∠BCP+∠BPC)
=180°-∠ABC
=120°
再问： 那31题了！谢谢
再答： (1): m^2=2n+1……① 4n^2=m+1……② ①-② m^2-4n^2=2n-m (m+2n)(m-2n)=2n-m m≠2n ∴m+2n=-1 (2): (m+2n)^2=(-1)^2=1 m^2+4mn+4n^2=1 2n+1+4mn+m+1=1 m+2n+4mn+1=0 -1+4mn+1=0 4mn=0 mn=0 ∴4n^3-mn+2n^2 =n(4n^2-m+2n) =n(m+1-m+2n) =n(2n+1) =nm^2 =nm×m =0×m =0