Tính:
M= \(\left(\frac{1}{2}-1\right).\left(\frac{1}{3}-1\right).\left(\frac{1}{4}-1\right)....\left(\frac{1}{2011}\right)\)
Những câu hỏi liên quan
\(\frac{1}{2011}.x=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{2010}\right).\left(1-\frac{1}{2011}\right)\)
\(\frac{1}{2011}.x=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{2010}\right).\left(1-\frac{1}{2011}\right)\)
\(\frac{1}{2011}.x=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{2009}{2010}.\frac{2010}{2011}\)
\(\frac{1}{2011}.x=\frac{1.2.3...2009.2010}{2.3.4...2010.2011}\)\(=\frac{1}{2011}\)
\(x=\frac{1}{2011}:\frac{1}{2011}=1\)
Vậy x=1
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\(\frac{1}{2011}.x=\frac{1}{2}.\left(\frac{2}{3}\right).\left(\frac{3}{4}\right)......\left(\frac{2010}{2011}\right)\)
\(\frac{1}{2011}.x=\frac{2}{4}.\left(\frac{4}{6}\right).\left(\frac{6}{8}\right).......\left(\frac{4018}{4020}\right).\left(\frac{4020}{4022}\right)\)
\(\frac{1}{2011}.x=\frac{2.4.6.8.....4018.4020}{4.6.8.10.....4020.4022}\)
\(\frac{1}{2011}.x=\frac{2}{4022}\)
\(\Rightarrow\)\(x=\frac{2}{4022}:\frac{1}{2011}=1\)
Ai thấy đún thì ủng hộ mink nha !!!
Thanks you very much !!
Chúc các bạn luôn học giỏi !!!
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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
\(B=\frac{6}{2}:\left(-1\frac{1}{2}\right):1\frac{1}{3}:\left(-1\frac{1}{4}\right):1\frac{1}{5}:\left(-1\frac{1}{6}\right):...:1\frac{1}{2011}:\left(-1\frac{1}{2012}\right)\)
\(C=\left(1+\frac{2}{3}\right).\left(1+\frac{2}{5}\right).\left(1+\frac{2}{7}\right).....\left(1+\frac{2}{2009}\right)+\left(1+\frac{2}{2011}\right)\)
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
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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
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(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
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a, Tính : frac{left(13frac{1}{4}-2frac{5}{27}-10frac{5}{6}right).230frac{1}{25}+46frac{3}{4}}{left(1frac{3}{10}+frac{10}{3}right):left(12frac{1}{3}-14frac{2}{7}right)}b, Tính : Pfrac{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+100right)left(frac{1}{2}-frac{1}{3}-frac{1}{7}-frac{1}{9}right)left(63.1,2-21.3,6right)}{1-2+3-4+...+99-100}
Đọc tiếp
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 : \(\frac{\left(1+2+3+...+99+100\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right)\left(63.1,2-21.3,6\right)}{1-2+3-4+...+99-100}\)
\(\left(1-\frac{1}{1+2}\right).\left(1-\frac{1}{1+2+3}\right)......\left(1-\frac{1}{1+2+3+4+....+2011}\right)\)
=\(\frac{2}{1+2}.\frac{2+3}{1+2+3}.\frac{2+3+4}{1+2+3+4}...\frac{2+3+4+...+2011}{1+2+3+....+2011}\)
=\(\frac{2}{\frac{\left(2+1\right).2}{2}}.\frac{\left(2+3\right).2}{\frac{2}{\frac{\left(3+1\right).3}{2}}}....\frac{\left(2+2011\right)\left(2011-1\right)}{\frac{2}{\frac{\left(2011+1\right)2011}{2}}}\)
=\(\frac{4}{\left(2+1\right).2}\frac{\left(2+3\right).2}{\left(3+1\right).3}....\frac{(2+2011)\left(2011-1\right)}{\left(2011+1\right)2011}\)
=\(\frac{\left(1.4\right)\left(5.2\right)....\left(2013.2010\right)}{\left(3.2\right).\left(4.3\right)....\left(2012.2011\right)}\)
=\(\frac{\left(1.2.3...2010\right)\left(4.5.6...2013\right)}{\left(2.3.4...2011\right)\left(3.4.5....2012\right)}\)
=\(\frac{1}{2011}.\frac{2013}{3}\)=\(\frac{671}{2011}\)
Mk nghĩ vậy. Chắc là đúng đấy
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Tính:
M= \(\left(\frac{1}{2}-1\right).\left(\frac{1}{3}-1\right).\left(\frac{1}{4}-1\right)...\left(\frac{1}{2011}-1\right)\)
\(M=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)...\left(\frac{1}{2011}-1\right)\)
\(M=-\frac{1}{2}.-\frac{2}{3}.-\frac{3}{4}...-\frac{2010}{2011}\)
\(M=-\frac{1}{2011}\)
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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 )
\(\left(1-\frac{2}{2.3}\right).\left(1-\frac{2}{3.4}\right)...\left(1-\frac{1}{2011.2012}\right)=\frac{4}{2.3}.\frac{10}{3.4}.\frac{18}{4.5}...\frac{2010.2013}{2011\cdot2012}\)
\(\frac{\left(1.4\right)\left(2.5\right)\left(3.6\right)...\left(2010.2013\right)}{\left(2.3.4...2011\right).\left(3.4.5....2012\right)}=\frac{\left(1.2.3...2010\right).\left(4.5.6....2013\right)}{\left(2.3.4.....2011\right)\left(3.4.5...2012\right)}=\frac{1.2013}{2011.3}\)
\(\frac{2013}{6033}\)
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\(\frac{1.2.3....2010}{2.3.4.....2011}\)= \(\frac{1}{2011}\)
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