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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 6: (1- 1/1+2). (1-1/1+2+3) . (1-1/1+2+3+4) -. ( 1 – 1/1+2+3+…+1996)

Toán Lớp 6: (1- 1/1+2). (1-1/1+2+3) . (1-1/1+2+3+4) ….. ( 1 – 1/1+2+3+…+1996)

Comments ( 2 )

  1. $Giải đáp+Lời giải và giải thích chi tiết:$
    $A=(1- 1/1+2).(1-1/1+2+3).(1-1/1+2+3+4) ….. ( 1 – 1/1+2+3+…+1996)$
    $=2/3.5/6.9/10….1993005/1993006$
    $=4/6.10/12.18/20…3986010/3986012$
    $=(1.4/2.3).(2.5/3.4).(3.6/4.5)…..(1995.1998/1996.1997)$
    $=(1.2.3…1995).(4.5…1998)/(2.3.4…1996).(3.4.5…1997)$
    $=1.1998/1996.3$
    $=333/998$
    $Vậy$ $A =333/998$

  2. Giải đáp:
    $\dfrac{333}{998}$
    Lời giải và giải thích chi tiết:
    $\left(1-\dfrac{1}{1+2}\right)\!.\!\left(1-\dfrac{1}{1+2+3}\right)\!.\!\left(1-\dfrac{1}{1+2+3+4}\right)\!.\!.\!.\!\left(1-\dfrac{1}{1+2+3\ +\,.\!.\!.+\ 1996}\right)\\=\left(1-\dfrac{1}{3}\right)\!.\!\left(1-\dfrac{1}{6}\right)\!.\!\left(1-\dfrac{1}{10}\right)\!.\!.\!.\!\bigg(1-\dfrac{1}{\dfrac{(1996+1).1996}{2}}\bigg)\\=\dfrac23\cdot\dfrac56\cdot\dfrac9{10}\ \cdot\,.\!.\!.\cdot\left(1-\dfrac{2}{1996.1997}\right)\\=\dfrac46\cdot\dfrac{10}{12}\cdot\dfrac{18}{20}\ \cdot\,.\!.\!.\cdot\ \dfrac{1996.1997-2}{1996.1997}\\=\dfrac46\cdot\dfrac{10}{12}\cdot\dfrac{18}{20}\ \cdot\,.\!.\!.\cdot\ \dfrac{(1995+1).1997-2}{1996.1997}\\=\dfrac46\cdot\dfrac{10}{12}\cdot\dfrac{18}{20}\ \cdot\,.\!.\!.\cdot\ \dfrac{1995.1997+1997-2}{1996.1997}\\=\dfrac46\cdot\dfrac{10}{12}\cdot\dfrac{18}{20}\ \cdot\,.\!.\!.\cdot\ \dfrac{1995.1997+1995}{1996.1997}\\=\dfrac46\cdot\dfrac{10}{12}\cdot\dfrac{18}{20}\ \cdot\,.\!.\!.\cdot\ \dfrac{1995.(1997+1)}{1996.1997}\\=\dfrac46\cdot\dfrac{10}{12}\cdot\dfrac{18}{20}\ \cdot\,.\!.\!.\cdot\ \dfrac{1995.1998}{1996.1997}\\=\dfrac{1.4}{2.3}\cdot\dfrac{2.5}{3.4}\cdot\dfrac{3.6}{4.5}\ \cdot\,.\!.\!.\cdot\ \dfrac{1995.1998}{1996.1997}\\=\dfrac{1.4.2.5.3.6…1995.1998}{2.3.3.4.4.5…1996.1997}\\=\dfrac{(1.2.3…1995).(4.5.6…1998)}{(2.3.4…1996).(3.4.5…1997)}\\=\dfrac{1.1998}{1996.3}\\=\dfrac{333}{998}$

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222-9+11+12:2*14+14 = ? ( )