Bài toán này giống của lớp 7 ghê

lớp 6 đó

1001

Yêu cầu bài toán chỉ đơn thuần tính cái này thôi à em!

Ta có: D\(=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{2005}\right)\)

\(\Leftrightarrow D=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{2004}{2005}=\dfrac{1.2.3...2004}{2.3.4...2005}=\dfrac{1}{2005}\)

Ta có: \(E=\dfrac{1^2}{1.3}.\dfrac{2^2}{2.4}.\dfrac{3^2}{3.5}...\dfrac{999^2}{999.1000}.\dfrac{1000^2}{1000.1001}=\dfrac{\left(1.2.3.4...1000\right)\left(1.2.3.4...1000\right)}{\left(1.2.3....1000\right)\left(3.4.5....1001\right)}=\dfrac{2}{1001}\)

\(...=\dfrac{3}{2}x\dfrac{4}{3}x\dfrac{5}{4}x\dfrac{6}{5}x\dfrac{7}{6}....x\dfrac{1000}{999}\)

\(=\dfrac{1}{2}x\dfrac{1000}{1}=500\)

=3/2x4/3x5/4x....x1000/999

=1/2x1000=500

mình chưa chắc là đúng đâu nhé

Ta có \(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}=\dfrac{1}{6}-\dfrac{1}{6}=0\) nên Q = 0.

\(Q=\left(\dfrac{1}{99}+\dfrac{12}{999}+\dfrac{123}{9999}\right).\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\) 

\(Q=\left(\dfrac{1}{99}+\dfrac{12}{999}+\dfrac{123}{9999}\right).0\) 

\(Q=0\)

Q=...
có thấy đa thức Q ghi j đâu

A = ( \(\dfrac{1}{2}\) + 1)(\(\dfrac{1}{3}\) + 1).(\(\dfrac{1}{4}\) + 1)....(\(\dfrac{1}{999}\) + 1)

A = \(\dfrac{3}{2}\).\(\dfrac{4}{3}\).\(\dfrac{5}{4}\).......\(\dfrac{1000}{999}\)

A = \(\dfrac{3.4.5......999}{3.4.5......999}\). \(\dfrac{1000}{2}\)

A = 1 \(\times\) 500

A = 500