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$a)\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{90}$
$=\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}$
$=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}$
$=\dfrac{1}{2}-\dfrac{1}{10}=\dfrac{2}{5}$
b) Đặt $B=\dfrac{4}{2.4}+\dfrac{4}{4.6}+...+\dfrac{4}{18.20}$
$=>\dfrac{1}{2}B=\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{18.20}$
$=>\dfrac{1}{2}B=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{18}-\dfrac{1}{20}$
$=>\dfrac{1}{2}B=\dfrac{1}{2}-\dfrac{1}{20}=\dfrac{9}{20}$
$=>B=\dfrac{9}{10}$
c) $\dfrac{2}{3}+\dfrac{2}{15}+\dfrac{2}{35}+\dfrac{2}{63}$
$=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}$
$=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}$
$=1-\dfrac{1}{9}=\dfrac{8}{9}$
d) Viết lại đề rõ ràng nha bạn.
Ta có: \(F=\dfrac{4}{2\cdot4}+\dfrac{4}{4\cdot6}+\dfrac{4}{6\cdot8}+...+\dfrac{4}{2008\cdot2010}\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)
\(=2\cdot\left(\dfrac{1}{2}-\dfrac{1}{2010}\right)\)
\(=2\cdot\dfrac{502}{1005}=\dfrac{1004}{1005}\)
\(F=\dfrac{4}{2.4}+\dfrac{4}{4.6}+\dfrac{4}{6.8}+...+\dfrac{4}{2008.2010}\)
\(F=2.\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{2008.2010}\right)\)
\(F=2.\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)
\(F=2.\left(\dfrac{1}{2}-\dfrac{1}{2010}\right)\)
\(F=1-\dfrac{1}{1005}=\dfrac{1004}{1005}\)
\(N=\dfrac{4}{2.4}+\dfrac{4}{4.6}+\dfrac{4}{6.8}+...+\dfrac{4}{2014.2016}\)
\(=2\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{2014.2016}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{2014}-\dfrac{1}{2016}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{2016}\right)\)
\(=2\left(\dfrac{1008}{2016}-\dfrac{1}{2016}\right)\)
\(=2.\dfrac{1007}{2016}=\dfrac{1007}{1008}\)
Công thức đây bạn:
\(\dfrac{a}{n\left(n+a\right)}=\dfrac{1}{n}-\dfrac{1}{n+a}\)
K = 2( 2/2.4 + 2/4.6 +......+ 2/2008.2010)
K = 2( 1/2 - 1/4 + 1/4 - 1/6 +......+ 1/2008 - 1/2010)
K = 2( 1/2 - 1/2010)
K = 2 . 1004/2010
K = 1004/1005
Ai k mk mk k lại
K=\(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
K=\(\frac{4}{2}.\frac{2}{2.4}+\frac{4}{2}.\frac{2}{4.6}+...+\frac{4}{2}.\frac{2}{2008.2010}\)
K=\(\frac{4}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2008.2010}\right)\)
K=\(\frac{4}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+..+\frac{1}{2008}-\frac{1}{2010}\right)\)
K=\(\frac{4}{2}.\left(\frac{1}{2}-\frac{1}{2010}\right)\)
K=\(\frac{4}{2}.\frac{502}{1005}\)
K=\(\frac{1004}{1005}\)
\(K=2\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2008.2010}\right)\)
\(K=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(K=2\left(\frac{1}{2}-\frac{1}{2010}\right)\)
\(K=2\times\frac{502}{1005}\)
\(K=\frac{1004}{1005}\)
Ta có : D = \(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+.....+\frac{4}{2008.2010}\)
\(\Leftrightarrow D=2\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+....+\frac{2}{2008.2010}\right)\)
\(\Leftrightarrow D=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(\Leftrightarrow D=2\left(\frac{1}{2}-\frac{1}{2010}\right)\)
\(\Leftrightarrow D=1-\frac{1}{1005}=\frac{1004}{1005}\)
D = 2.(2/2.4+2/4.6+...+2/2008.2010)
=2(1/2-1/4+1/4-1/6+......+1/2008-1/2
=2(1/2-1/2010)
=2.502/1005
=1004/1005
A=3n+1/n-1=3(n-1)+4/n-1=3+4/n-1
Để A là số nguyên thì 4/n-1 là số nguyên
=>n-1 thuộc Ư(4)=1,-1,2,-2,4,-4
=>n thuộc (2,0,3,-1,5,-3)
Ta có : \(A=\frac{3n+2}{n-1}+\frac{3n-3+5}{n-1}=\frac{3\left(n-1\right)+5}{n-1}=\frac{3\left(n-1\right)}{n-1}+\frac{5}{n-1}=3+\frac{5}{n-1}\)
Để A có giá trị nguyên thì n - 1 thuộc Ư(5) = {-1;-5;1;5}
| n - 1 | -5 | -1 | 1 | 5 |
| n | -4 | 0 | 2 | 6 |
| A = \(3+\frac{5}{n-1}\) | 2 | -2 | 8 | 4 |
\(F=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
\(F=2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(F=2.\left(\frac{1}{2}-\frac{1}{2010}\right)\)
\(F=2.\frac{502}{1005}\)
\(F=\frac{1004}{1005}\)
nhinf vào là biết luật ngay bài đó bằng = \(\frac{1004}{1005}\)
kết bạn với mình nha
A=4/2.4+4/4.6+4/6.8+...+4/2008.2010
=2.(2/2.4+2/4.6+2/6.8+...+2/2008.2010)
=2.(1/2-1/4+1/4-1/6+1/6-1/8+...+1/2008-1/2010)
=2.(1/2-1/2010)
=2.502/1005
=1004/1005
Vậy A=1004/1005
100% giải đúng đầu tiên:
Ta có: \(A=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
\(=2.\frac{2}{2.4}+2.\frac{2}{4.6}+2.\frac{2}{6.8}+...+2.\frac{2}{2008.2010}\)
\(=2\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+..+\frac{2}{2008.2010}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{2010}\right)\)
\(=2.\frac{1}{2}-2.\frac{1}{2010}\)
\(=1-\frac{1}{1005}=\frac{1004}{1005}\)
\(\dfrac{4}{2\cdot4}+\dfrac{4}{4\cdot6}+\dfrac{4}{6\cdot8}+...+\dfrac{4}{16\cdot18}+\dfrac{4}{18\cdot20}\)
\(=2\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{16\cdot18}+\dfrac{2}{18\cdot20}\right)\)
\(=2\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{16}-\dfrac{1}{18}+\dfrac{1}{18}-\dfrac{1}{20}\right)\)
\(=2\left(1-\dfrac{1}{20}\right)\)
\(=2\left(\dfrac{20}{20}-\dfrac{1}{20}\right)\)
\(=2\cdot\dfrac{19}{20}\)
\(=\dfrac{19}{10}\)
\(\dfrac{4}{2.4}+\dfrac{4}{4.6}+\dfrac{4}{6.8}+.....+\dfrac{4}{16.18}+\dfrac{4}{18.20}\)
\(=\dfrac{4}{2}\left(\dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+.....+\dfrac{1}{16.18}+\dfrac{1}{18.20}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+.....+\dfrac{1}{16}-\dfrac{1}{18}+\dfrac{1}{18}-\dfrac{1}{20}\right)\)\(=2\left(1-\dfrac{1}{20}\right)=2.\dfrac{19}{20}=\dfrac{19}{10}\)