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\(\frac{1}{2011}.x=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{2010}\right).\left(1-\frac{1}{2011}\right)\)
\(\frac{1}{2011}.x=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{2009}{2010}.\frac{2010}{2011}\)
\(\frac{1}{2011}.x=\frac{1.2.3...2009.2010}{2.3.4...2010.2011}\)\(=\frac{1}{2011}\)
\(x=\frac{1}{2011}:\frac{1}{2011}=1\)
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 !!!
\(M=\left[\frac{1}{2}-1\right]\cdot\left[\frac{1}{3}-1\right]\cdot\left[\frac{1}{4}-1\right]\cdot...\cdot\left[\frac{1}{2011}-1\right]\)
\(M=\left[\frac{1}{2}-\frac{2}{2}\right]\cdot\left[\frac{1}{3}-\frac{3}{3}\right]\cdot\left[\frac{1}{4}-\frac{4}{4}\right]\cdot...\cdot\left[\frac{1}{2011}-\frac{2011}{2011}\right]\)
\(M=\frac{-1}{2}\cdot\frac{-2}{3}\cdot\frac{-3}{4}\cdot...\cdot\frac{-2010}{2011}\)
\(M=\frac{\left[-1\right]\cdot\left[-2\right]\cdot\left[-3\right]\cdot...\cdot\left[-2010\right]}{2\cdot3\cdot4\cdot...\cdot2011}\)
\(M=\frac{1\cdot2\cdot3\cdot...\cdot2010}{2\cdot3\cdot4\cdot...\cdot2011}=\frac{1}{2011}\)
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
C=(1+2/3).(1+2/5).(1+2/7)......(1+2/2009).(1+2/2011)
C=5/3.7/5.9/7......2011/2009.2013/2011
C=5.7.9.....2013/3.5.7.....2009.2011
C=2013/3
trong dãy tích A sẽ có phân số \(1-\frac{2010}{2010}=1-1=0\)
=>A=0
\(A=\frac{ }{ }sdadsad\text{đ}\text{s}gh\text{d}fg\text{d}\)sf
\(1.\)\(M=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{42}\)
\(M=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{6.7}\)
\(M=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{6}-\frac{1}{7}\)
\(M=1-\frac{1}{7}=\frac{6}{7}\)
Mình làm câu 1 thoi nha!
1.
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)
=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)
=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{6}-\frac{1}{7}\)
=\(1-\frac{1}{7}\)
=\(\frac{6}{7}\)
\(M=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)...\left(\frac{1}{2011}-1\right)\)
\(M=-\frac{1}{2}.-\frac{2}{3}.-\frac{3}{4}...-\frac{2010}{2011}\)
\(M=-\frac{1}{2011}\)
M =-1/2011