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\(\frac{1}{2}.4+\frac{1}{4}.6+\frac{1}{6}.8+\frac{1}{8}.10\)
=\(\frac{73}{12}\)
k mình nha
18:
a: \(S=3\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{98\cdot100}\right)\)
=3*(1/2-1/4+1/4-1/6+...+1/98-1/100)
=3*49/100=147/100
b: Để A là số nguyên thì n-1 thuộc Ư(2)
=>n-1 thuộc {1;-1;2;-2}
=>n thuộc {2;0;3;-1}
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: \(A=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
\(=2.\frac{2}{2.4}+2.\frac{2}{4.6}+2.\frac{2}{6.8}+...+2.\frac{2}{2008.2010}\)
\(=2\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+..+\frac{2}{2008.2010}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{2010}\right)\)
\(=2.\frac{1}{2}-2.\frac{1}{2010}\)
\(=1-\frac{1}{1005}=\frac{1004}{1005}\)
a: \(=\dfrac{3\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}{4\left(\dfrac{1}{47}-\dfrac{4}{47}+\dfrac{9}{53}\right)}=\dfrac{3}{4}\)
b: \(F=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)
\(=2\cdot\dfrac{1004}{2010}=\dfrac{2008}{2010}=\dfrac{1004}{1005}\)
c: \(S=\dfrac{1}{3\cdot6}+\dfrac{1}{6\cdot9}+...+\dfrac{1}{30\cdot33}\)
\(=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+...+\dfrac{1}{30}-\dfrac{1}{33}\right)\)
\(=\dfrac{1}{3}\cdot\dfrac{10}{33}=\dfrac{10}{99}\)
=> K : 2 = \(\frac{2}{2.4}+\frac{2}{4.6}+....+\frac{2}{2008.2010}\)
= \(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2008}-\frac{1}{2010}\)
=\(\frac{1}{2}-\frac{1}{2010}=\frac{502}{1005}\)
\(\Rightarrow K=\frac{1004}{1005}\)
Vậy \(K=\frac{1004}{1005}\)
F=2 .(1/2-1/4+1/4-1/6+......+1/2008 - 1/2010)
= 2.(1/2-1/2010)
= 2. 502/1005
= 1004/1005
Ta có: \(F=\dfrac{4}{2\cdot4}+\dfrac{4}{4\cdot6}+\dfrac{4}{6\cdot8}+...+\dfrac{4}{2008\cdot2010}\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)
\(=2\cdot\left(\dfrac{1}{2}-\dfrac{1}{2010}\right)\)
\(=2\cdot\dfrac{502}{1005}=\dfrac{1004}{1005}\)
\(F=\dfrac{4}{2.4}+\dfrac{4}{4.6}+\dfrac{4}{6.8}+...+\dfrac{4}{2008.2010}\)
\(F=2.\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{2008.2010}\right)\)
\(F=2.\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)
\(F=2.\left(\dfrac{1}{2}-\dfrac{1}{2010}\right)\)
\(F=1-\dfrac{1}{1005}=\dfrac{1004}{1005}\)
S = \(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2016.2018}\)
S = \(2.\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2016.2018}\right)\)
S = \(2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2016}-\frac{1}{2018}\right)\)
S = \(2.\left(\frac{1}{2}-\frac{1}{2018}\right)\)
S = \(2.\frac{504}{1009}\)= \(\frac{1008}{1009}\)
Vậy S = \(\frac{1008}{1009}\).
~~~
Nếu có sai sót gì thì giúp đỡ tớ nha :3
#Sunrise
\(F=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
\(F=2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(F=2.\left(\frac{1}{2}-\frac{1}{2010}\right)\)
\(F=2.\frac{502}{1005}\)
\(F=\frac{1004}{1005}\)
nhinf vào là biết luật ngay bài đó bằng = \(\frac{1004}{1005}\)
kết bạn với mình nha