Toán Lớp 6: Cho A = 1/1.99 + 1/2.98 +…+1/99.1
B = 1+ 1/2 + 1/3 +…+1/99
Tính B/A
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Toán Lớp 6: Cho A = 1/1.99 + 1/2.98 +…+1/99.1
B = 1+ 1/2 + 1/3 +…+1/99
Tính B/A
Comments ( 2 )
Giải đáp:
$\dfrac BA=50$.
Lời giải và giải thích chi tiết:
$\dfrac BA=\dfrac{1+\dfrac12+\dfrac13\ +\,.\!.\!.+\ \dfrac1{97}+\dfrac1{98}+\dfrac1{99}}{\dfrac1{1.99}+\dfrac1{2.98}\ +\,.\!.\!.+\ \dfrac1{98.2}+\dfrac1{99.1}}\\\Rightarrow \dfrac BA=\dfrac{\left(1+\dfrac1{99}\right)\!+\!\left(\dfrac12+\dfrac1{98}\right)\!+\!\left(\dfrac13+\dfrac1{97}\right) +\,.\!.\!.}{\left(\dfrac1{1.99}+\dfrac1{99.1}\right)\!+\!\left(\dfrac1{2.98}+\dfrac1{98.2}\right)\!+\!\left(\dfrac1{3.97}+\dfrac1{97.3}\right)\,.\!.\!.}\\\Rightarrow \dfrac BA=\dfrac{\dfrac{100}{1.99}+\dfrac{100}{2.98}+\dfrac{100}{3.97}\ +\,.\!.\!.}{\dfrac2{1.99}+\dfrac2{2.98}+\dfrac2{3.97}\ +\,.\!.\!.}\\\Rightarrow\dfrac BA=\dfrac{100.\!\left(\dfrac1{1.99}+\dfrac1{2.98}+\dfrac1{3.97}\ +\,.\!.\!.\right)}{\ \ \ \ 2.\!\left(\dfrac1{1.99}+\dfrac1{2.98}+\dfrac1{3.97}\ +\,.\!.\!.\right)}\\\Rightarrow \dfrac BA=\dfrac{100}2\\\Rightarrow\dfrac BA=50$
Vậy $\dfrac BA=50$.
A={99/1+98/2+…+1/99}/{1+1/2+1/3+…+1/99}
<=>A=99/1 + 98/2 +…+1/99 = (98/2 + 1) + (97/3 + 1) +…+(1/99 + 1) +99- 99
<=>A=100/2 +100/3 +…+100/99 = 100. (1/2 +1/3 +…+1/99)
=> A = {100. (1/2 +1/3 +…+1/99)}/{(1/2 +1/3 +..1/99)} =100
B=1+1/2+1/3+…+1/99
B=1+1/2+1/3+1/4+…….+1/99
B=2-1+1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+……+1/98-1/99
B=2-1/99
B=197/99