a)\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{5}\right)...\left(1-\frac{1}{20}\right)\)

\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{19}{20}\)

\(A=\frac{1.2.3...19}{2.3.4...20}\)

\(A=\frac{1}{20}\)

\(\frac{1}{2011}.x=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{2010}\right).\left(1-\frac{1}{2011}\right)\)

\(\frac{1}{2011}.x=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{2009}{2010}.\frac{2010}{2011}\)

\(\frac{1}{2011}.x=\frac{1.2.3...2009.2010}{2.3.4...2010.2011}\)\(=\frac{1}{2011}\)

\(x=\frac{1}{2011}:\frac{1}{2011}=1\)

Vậy x=1

\(\frac{1}{2011}.x=\frac{1}{2}.\left(\frac{2}{3}\right).\left(\frac{3}{4}\right)......\left(\frac{2010}{2011}\right)\)

\(\frac{1}{2011}.x=\frac{2}{4}.\left(\frac{4}{6}\right).\left(\frac{6}{8}\right).......\left(\frac{4018}{4020}\right).\left(\frac{4020}{4022}\right)\)

\(\frac{1}{2011}.x=\frac{2.4.6.8.....4018.4020}{4.6.8.10.....4020.4022}\)

\(\frac{1}{2011}.x=\frac{2}{4022}\)

\(\Rightarrow\)\(x=\frac{2}{4022}:\frac{1}{2011}=1\)

Ai thấy đún thì ủng hộ mink nha !!!

Thanks you very much !!

Chúc các bạn luôn học giỏi !!!

Số cuối có trừ 2 koe

\(M=\left[\frac{1}{2}-1\right]\cdot\left[\frac{1}{3}-1\right]\cdot\left[\frac{1}{4}-1\right]\cdot...\cdot\left[\frac{1}{2011}-1\right]\)

\(M=\left[\frac{1}{2}-\frac{2}{2}\right]\cdot\left[\frac{1}{3}-\frac{3}{3}\right]\cdot\left[\frac{1}{4}-\frac{4}{4}\right]\cdot...\cdot\left[\frac{1}{2011}-\frac{2011}{2011}\right]\)

\(M=\frac{-1}{2}\cdot\frac{-2}{3}\cdot\frac{-3}{4}\cdot...\cdot\frac{-2010}{2011}\)

\(M=\frac{\left[-1\right]\cdot\left[-2\right]\cdot\left[-3\right]\cdot...\cdot\left[-2010\right]}{2\cdot3\cdot4\cdot...\cdot2011}\)

\(M=\frac{1\cdot2\cdot3\cdot...\cdot2010}{2\cdot3\cdot4\cdot...\cdot2011}=\frac{1}{2011}\)

M= (1/2 -1).(1/3 - 1).(1/4 - 1)...(1/2011)

M= -1/2 . -2/3 . -3/4 ... 1/2011

M= -1 . -2 . -3 ...1 / 2 . 3 .4 . 2011

M= -1/2011

B

từ 1 đến 2012 có tất cả:

2012-1:1+1 = 2012 (số)

=>có: 2012:2 = 1006 (cặp)

Mà mỗi cặp bằng (-1)nên

tổng dãy số trên là: 1006 . (-1) = -1006

(1-2)+(2-3)+(3-4)+(5-6)+...+(2011-2012)

=-1+(-1)+(-1)+(-1)+...+(-1)

có tất cả các số -1 trên dãy số trên là

(2012-2);2+1=1006

vậy suy ra ; -1x1006=(-1006)

chac chan la dung

quá dễ tách ra thành 1\x-1\x+1+1\x+1-1\x+2+1\x+2-1\x+3+1\x+3-1\x+4+...+1\x+5-1\x+6

=1\x-1\x+6

=6\x(x+6)

\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}\)\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}\)

\(=\frac{1}{x}-\frac{1}{x+6}\)\(=\frac{6}{x\left(x+6\right)}\)

Tổng các số tự nhiên từ 1 đến n là \(\frac{n\left(n+1\right)}{2}\)

Do đó \(A=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+....+\frac{1}{2011}.\frac{2011.2012}{2}\)

\(=1+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{2012}{2}\)

\(=\left(\frac{1}{2}+\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+...+\frac{2012}{2}\right)-\frac{1}{2}\)

\(=\frac{1+2+3+...+2012}{2}-\frac{1}{2}\)

\(=\frac{\frac{2012.2013}{2}}{2}-\frac{1}{2}\)

\(=1012538,5\)

Vậy ....

A=(n+1)(n+2)/4=2012.2013/4=503.2013