\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3.4}+\frac{1}{4.5}=x\)
Tìm x biết dấu chấm là dấu nhân.
K
Khách
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Những câu hỏi liên quan
12 tháng 6 2018
\(a,\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2017\cdot2018}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(=1-\frac{1}{2018}\)
\(=\frac{2017}{2018}.\)
\(b,\left[x\cdot\frac{5}{3}-1\right]:9=3\frac{1}{2}:2,25\)
\(\Leftrightarrow\left[x\cdot\frac{5}{3}-1\right]:9=\frac{7}{2}:\frac{9}{4}\)
\(\Leftrightarrow\left[x\cdot\frac{5}{3}-1\right]:9=\frac{7}{2}\cdot\frac{4}{9}\)
\(\Leftrightarrow\left[x\cdot\frac{5}{3}-1\right]:9=\frac{14}{9}\)
\(\Leftrightarrow x\cdot\frac{5}{3}-1=\frac{14}{9}\cdot9\)
\(\Leftrightarrow x\cdot\frac{5}{3}-1=14\)
\(\Leftrightarrow x\cdot\frac{5}{3}=14+1\)
\(\Leftrightarrow x\cdot\frac{5}{3}=15\)
\(\Leftrightarrow x=15:\frac{5}{3}\)
\(\Leftrightarrow x=15\cdot\frac{3}{5}\)
\(\Leftrightarrow x=9.\)
12 tháng 6 2018
a)\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2017.2018}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(=\frac{1}{1}-\frac{1}{2018}\)
\(=\frac{2017}{2018}\)
b)\(\left[x.\frac{5}{3}-1\right]:9=3\frac{1}{2}:2,25\)
\(\Leftrightarrow\left[x.\frac{5}{3}-1\right]:9=3\frac{1}{2}:\frac{9}{4}=1\frac{5}{9}\)
\(\Rightarrow x.\frac{5}{3}-1=1\frac{5}{9}.9=14\)
\(\Rightarrow x.\frac{5}{3}=14+1=15\)
\(\Rightarrow x=15:\frac{5}{3}=9\)
1 tháng 7 2016
1/1x2+1/2x3+1/3x4+...+1/99x100+1
1/1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100 +1
=1- 1/100 +1
=99/100 +1
=199/100
1 tháng 7 2016
1/1.2 + 1/2.3 + 1/3.4 + ... + 1/999.1000 + 1
= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/999 - 1/1000 + 1
= 1 - 1/1000 + 1
= 2 - 1/1000
= 2000/1000 - 1/1000
= 1999/1000
Ủng hộ mk nha ♡_♡☆_☆
HN
1 tháng 7 2018
a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.......+\frac{1}{2017.2018}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-..........-\frac{1}{2018}\)
\(=1-\frac{1}{2018}\)
\(=\frac{2018}{2018}-\frac{1}{2018}=\frac{2017}{2018}\)
b) \(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+..........+\frac{2}{2017.2018}+\frac{2}{2018.2019}\)
\(=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.........+\frac{1}{2017.2018}+\frac{1}{2018.2019}\right)\)
\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-.........-\frac{1}{2018}+\frac{1}{2018}-\frac{1}{2019}\right)\)
\(=2\left(1-\frac{1}{2019}\right)\)
\(=2\left(\frac{2019}{2019}-\frac{1}{2019}\right)\)
\(=2.\frac{2018}{2019}\)
\(=\frac{4036}{2019}\)
Phần c tương tự nha
1 tháng 7 2018
a) \(\frac{1}{1.2}\) + \(\frac{1}{2.3}\) + .......+ \(\frac{1}{2017.2018}\)
= 1 - \(\frac{1}{2}\) + \(\frac{1}{2}\) - \(\frac{1}{3}\) + .......+ \(\frac{1}{2017}\) - \(\frac{1}{2018}\)
= 1 - \(\frac{1}{2018}\) = \(\frac{2017}{2018}\)
câu a) mik sửa đề một tí ko biết có đúng ko
câu b , c tương tự nhưng cần lấy tử ra chung
DP
23 tháng 7 2017
\(C=\frac{9}{10}-\frac{1}{10.9}-\frac{1}{9.8}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(C=\frac{9}{10}-\left(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{9.8}+\frac{1}{9.10}\right)\)
\(C=\frac{9}{10}-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
\(C=\frac{9}{10}-\left(\frac{1}{1}-\frac{1}{10}\right)\)
\(C=\frac{9}{10}-\frac{9}{10}=0\)
NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 3 2019
Ta có:
\(\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{2}\left(\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)\)
\(\Rightarrow\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{2005.2006.2007}=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+...+\frac{1}{2005.2006}-\frac{1}{2006.2007}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2006.2007}\right)=\frac{1}{2}\left(\frac{2005.2008}{2.2006.2007}\right)\)
Đặt \(A=1.2+2.3+...+n\left(n+1\right)\)
\(\Rightarrow3A=1.2.\left(3-0\right)+2.3.\left(4-1\right)+...+n\left(n+1\right)\left(n+2-\left(n-1\right)\right)\)
\(\Rightarrow3A=1.2.3-1.2.0+2.3.4-1.2.3+...+n\left(n+1\right)\left(n+2\right)-\left(n-1\right)n\left(n+1\right)\)
\(\Rightarrow3A=n\left(n+1\right)\left(n+2\right)\)
\(\Rightarrow A=\frac{n\left(n+1\right)\left(n+2\right)}{3}\)
\(\Rightarrow1.2+2.3+...+2006.2007=\frac{2006.2007.2008}{2}\)
Vậy pt trở thành:
\(\frac{1}{2}\left(\frac{2005.2008}{2.2006.2007}\right)x=\frac{2006.2007.2008}{2}\)
\(\Leftrightarrow\frac{2005}{2.2006.2007}x=2006.2007\)
\(\Rightarrow x=\frac{2.\left(2006.2007\right)^2}{2005}\)
KN
6 tháng 5 2018
Bài 1
a) \(P=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{10}{10}-\frac{1}{10}=\frac{9}{10}\)
b) \(S=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}\)
\(=\frac{33}{99}-\frac{1}{99}\)
\(=\frac{32}{99}\)
c)\(Q=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{19.20}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{20}\)
\(=\frac{1}{2}-\frac{1}{20}\)
\(=\frac{10}{20}-\frac{1}{20}\)
\(=\frac{9}{20}\)
Tk mình nha!!
KN
7 tháng 5 2018
Câu 2:
\(P=\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right).\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{99}\right)\)
\(=\left(\frac{2}{2}+\frac{1}{2}\right).\left(\frac{3}{3}+\frac{1}{3}\right).\left(\frac{4}{4}+\frac{1}{4}\right)...\left(\frac{99}{99}+\frac{1}{99}\right)\)
\(=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot...\cdot\frac{100}{99}\)
\(=\frac{3\cdot4\cdot5...100}{2.3.4...99}\)
\(=\frac{3\cdot100}{2}\)
\(=\frac{300}{2}=150\)
HN
5 tháng 7 2016
c.\(=3\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+..+\frac{2}{99.101}\right)\)
\(=3\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=3\left(1-\frac{1}{101}\right)\)
\(=\frac{300}{101}\)
2 tháng 8 2017
\(\frac{x}{6}+\frac{7}{3.2^2}=\frac{17}{18}-\frac{1}{3^2}\)
\(\frac{x}{6}+\frac{7}{12}=\frac{5}{6}\)
\(\frac{x}{6}=\frac{5}{6}-\frac{7}{12}\)
\(\frac{x}{6}=\frac{1}{4}\)
\(x=\frac{1}{4}.6\)
\(x=\frac{3}{2}\)
\(x=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}\)
\(x=\frac{1}{1}-\frac{1}{5}=\frac{4}{5}\)
x=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5
x=1-1/5
x=4/5