Tính:
M= \(\left(\frac{1}{2}-1\right).\left(\frac{1}{3}-1\right).\left(\frac{1}{4}-1\right)...\left(\frac{1}{2011}-1\right)\)
\(\frac{1}{2011}.x=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{2010}\right).\left(1-\frac{1}{2011}\right)\)
\(\frac{1}{2011}.x=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{2010}\right).\left(1-\frac{1}{2011}\right)\)
\(\frac{1}{2011}.x=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{2009}{2010}.\frac{2010}{2011}\)
\(\frac{1}{2011}.x=\frac{1.2.3...2009.2010}{2.3.4...2010.2011}\)\(=\frac{1}{2011}\)
\(x=\frac{1}{2011}:\frac{1}{2011}=1\)
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 !!!
Tính:
M= \(\left(\frac{1}{2}-1\right).\left(\frac{1}{3}-1\right).\left(\frac{1}{4}-1\right)....\left(\frac{1}{2011}\right)\)
\(M=\left[\frac{1}{2}-1\right]\cdot\left[\frac{1}{3}-1\right]\cdot\left[\frac{1}{4}-1\right]\cdot...\cdot\left[\frac{1}{2011}-1\right]\)
\(M=\left[\frac{1}{2}-\frac{2}{2}\right]\cdot\left[\frac{1}{3}-\frac{3}{3}\right]\cdot\left[\frac{1}{4}-\frac{4}{4}\right]\cdot...\cdot\left[\frac{1}{2011}-\frac{2011}{2011}\right]\)
\(M=\frac{-1}{2}\cdot\frac{-2}{3}\cdot\frac{-3}{4}\cdot...\cdot\frac{-2010}{2011}\)
\(M=\frac{\left[-1\right]\cdot\left[-2\right]\cdot\left[-3\right]\cdot...\cdot\left[-2010\right]}{2\cdot3\cdot4\cdot...\cdot2011}\)
\(M=\frac{1\cdot2\cdot3\cdot...\cdot2010}{2\cdot3\cdot4\cdot...\cdot2011}=\frac{1}{2011}\)
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
tính
\(B=\frac{6}{2}:\left(-1\frac{1}{2}\right):1\frac{1}{3}:\left(-1\frac{1}{4}\right):1\frac{1}{5}:\left(-1\frac{1}{6}\right):...:1\frac{1}{2011}:\left(-1\frac{1}{2012}\right)\)
a, Tính : \(\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right).230\frac{1}{25}+46\frac{3}{4}}{\left(1\frac{3}{10}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)
b, Tính : \(P=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{1}{2011}}\)
c, Tính : \(\frac{\left(1+2+3+...+99+100\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right)\left(63.1,2-21.3,6\right)}{1-2+3-4+...+99-100}\)
Tính Tổng : \(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).......\left(1-\frac{1}{2011}\right)\)
$\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).......\left(1-\frac{1}{2011}\right)$(1−12 )(1−13 )(1−14 ).......(1−12011 )
\(\left(1-\frac{2}{2.3}\right).\left(1-\frac{2}{3.4}\right)...\left(1-\frac{1}{2011.2012}\right)=\frac{4}{2.3}.\frac{10}{3.4}.\frac{18}{4.5}...\frac{2010.2013}{2011\cdot2012}\)
\(\frac{\left(1.4\right)\left(2.5\right)\left(3.6\right)...\left(2010.2013\right)}{\left(2.3.4...2011\right).\left(3.4.5....2012\right)}=\frac{\left(1.2.3...2010\right).\left(4.5.6....2013\right)}{\left(2.3.4.....2011\right)\left(3.4.5...2012\right)}=\frac{1.2013}{2011.3}\)
\(\frac{2013}{6033}\)
\(\frac{1.2.3....2010}{2.3.4.....2011}\)= \(\frac{1}{2011}\)
Thực hiện phép tính:
a) \(\left(\frac{1}{3}-1\right).\left(\frac{1}{6}-1\right).\left(\frac{1}{10}-1\right).\left(\frac{1}{15}-1\right)...\left(\frac{1}{1225}-1\right).\left(\frac{1}{1275}-1\right)\)
b) 1 - 2 + 3 - 4 + 5 - 6 + ... + 2011 - 2012
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
1.tính tổng
a. A=\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{20}\right)\)
b. B=\(\left(1-\frac{1}{2010}\right).\left(1-\frac{2}{2010}\right).\left(1-\frac{3}{2010}\right).....\left(1-\frac{2011}{2010}\right)\)
a)\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{5}\right)...\left(1-\frac{1}{20}\right)\)
\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{19}{20}\)
\(A=\frac{1.2.3...19}{2.3.4...20}\)
\(A=\frac{1}{20}\)
Tính ; A =1+\(\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{2011}\left(1+2+3+...+2011\right)\)
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 ....
Tính:
a.A = \(\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right).230\frac{1}{25}+46\frac{3}{4}}{\left(1\frac{3}{10}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)
b. B = \(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{1}{2011}}\)
c. C = \(\frac{\left(1+2+3+...+99+100\right).\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right).\left(63.1,2-21.3,6\right)}{1-2+3-4+...+99-100}\)