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)

TRẢ LỜI

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

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

Trả lời