cho biểu thức : A= 1/1×2 + 3/2×5+9/5×14+23/14×37+15/37×52+1967/42×2619 chững minh A<1 01/11/2021 Bởi Madeline cho biểu thức : A= 1/1×2 + 3/2×5+9/5×14+23/14×37+15/37×52+1967/42×2619 chững minh A<1
Nếu :$A=\dfrac{1}{1.2} + \dfrac{3}{2.5} + \dfrac{9}{5.14} + \dfrac{23}{14.37} + \dfrac{15}{37.52} + \dfrac{1967}{52.2019}$ $⇔ A = 1 – \dfrac{1}{2} +\dfrac{1}{2} – \dfrac{1}{5} + ………. + \dfrac{1}{52} – \dfrac{1}{2019}$ $⇔ A =1 – \dfrac{1}{2019}$ Vì $1 – \dfrac{1}{2019}$ < $1$ $⇒$ $A<1$ Vậy $A<1$($đpcm$) Nếu:$A=\dfrac{1}{1.2} + \dfrac{3}{2.5} + \dfrac{9}{5.14} + \dfrac{23}{14.37} + \dfrac{15}{37.52} + \dfrac{2567}{52.2619}$ $⇔ A = 1 – \dfrac{1}{2} +\dfrac{1}{2} – \dfrac{1}{5} + ………. + \dfrac{1}{52} – \dfrac{1}{2619}$ $⇔ A =1 – \dfrac{1}{2619}$ Vì $1 – \dfrac{1}{2169}$ < $1$ $⇒$ $A<1$ Vậy $A<1$($đpcm$) Bình luận
Nếu :$A=\dfrac{1}{1.2} + \dfrac{3}{2.5} + \dfrac{9}{5.14} + \dfrac{23}{14.37} + \dfrac{15}{37.52} + \dfrac{1967}{52.2019}$
$⇔ A = 1 – \dfrac{1}{2} +\dfrac{1}{2} – \dfrac{1}{5} + ………. + \dfrac{1}{52} – \dfrac{1}{2019}$
$⇔ A =1 – \dfrac{1}{2019}$
Vì $1 – \dfrac{1}{2019}$ < $1$ $⇒$ $A<1$
Vậy $A<1$($đpcm$)
Nếu:$A=\dfrac{1}{1.2} + \dfrac{3}{2.5} + \dfrac{9}{5.14} + \dfrac{23}{14.37} + \dfrac{15}{37.52} + \dfrac{2567}{52.2619}$
$⇔ A = 1 – \dfrac{1}{2} +\dfrac{1}{2} – \dfrac{1}{5} + ………. + \dfrac{1}{52} – \dfrac{1}{2619}$
$⇔ A =1 – \dfrac{1}{2619}$
Vì $1 – \dfrac{1}{2169}$ < $1$ $⇒$ $A<1$
Vậy $A<1$($đpcm$)