Tính: A= 2/1.4 + 2/4.7 + 2/7.10 + .....+ 2/196.199
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TQ
1
DD
8 tháng 5 2023
A = \(\dfrac{3^2}{1.4}+\dfrac{3^2}{4.7}+...+\dfrac{3^2}{196.199}\)
A = \(\dfrac{3.3}{1.4}+\dfrac{3.3}{4.7}+...+\dfrac{3.3}{196.199}\)
A = \(3.\dfrac{3}{1.4}+3.\dfrac{3}{4.7}+...+3.\dfrac{3}{196.199}\)
A = \(3\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{196.199}\right)\)
A = \(3\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{196}-\dfrac{1}{199}\right)\)
A = \(3\left(1-\dfrac{1}{199}\right)\) = \(3.\dfrac{198}{199}\) = \(\dfrac{594}{199}\)
TK
1
4 tháng 5 2017
\(A=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{97.100}\)
\(A=\frac{2}{3}.\left(1-\frac{1}{4}\right)+\frac{2}{3}.\left(\frac{1}{4}-\frac{1}{7}\right)+\frac{2}{3}.\left(\frac{1}{7}-\frac{1}{10}\right)+...+\frac{2}{3}.\left(\frac{1}{97}-\frac{1}{100}\right)\)
\(A=\frac{2}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(A=\frac{2}{3}.\left(1-\frac{1}{100}\right)\)
\(A=\frac{2}{3}.\frac{99}{100}\)
\(A=\frac{33}{50}\)
11 tháng 9 2016
Ta có: \(A=\frac{2}{1.4}+\frac{2}{4.7}+...+\frac{2}{97.100}\)
\(=\frac{2}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{97.100}\right)\)
Nhận xét: \(\frac{a}{x.\left(x+a\right)}=\frac{1}{x}-\frac{1}{x+a}\)
Do đó: \(A=\frac{2}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(=\frac{2}{3}.\left(1-\frac{1}{100}\right)\)
\(=\frac{2}{3}.\left(\frac{100}{100}-\frac{1}{100}\right)\)
\(=\frac{2}{3}.\frac{99}{100}\)
\(=\frac{33}{50}\)
Vậy,\(A=\frac{33}{50}\)
11 tháng 9 2016
\(\text{Ta có: }A=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+....+\frac{2}{97.100}\)
\(\Rightarrow\frac{3}{2}A=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{97.100}\)
\(\Rightarrow\frac{3}{2}A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{97}-\frac{1}{100}\)
\(\Rightarrow\frac{3}{2}A=1-\frac{1}{100}\)
\(\Rightarrow\frac{3}{2}A=\frac{99}{100}\)
\(\Rightarrow A=\frac{99}{100}:\frac{3}{2}\)
\(A=\frac{99}{100}.\frac{2}{3}=\frac{33}{50}\)
20 tháng 9 2016
\(A=\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+...+\frac{3^2}{97.100}\)
\(\Rightarrow A=3\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\right)\)
\(\Rightarrow A=3\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(\Rightarrow A=3\left(1-\frac{1}{100}\right)\)
\(\Rightarrow A=3.\frac{99}{100}\)
\(\Rightarrow A=3.\frac{99}{100}\)
\(\Rightarrow A=\frac{297}{100}\)
NT
1
14 tháng 2 2016
Mình mới học lớp 5 , xin lỗi nhé, mình cũng rất muốn giúp bạn nhưng ko đc.
18 tháng 9 2016
\(A=\frac{9}{1.4}+\frac{9}{4.7}+\frac{9}{7.10}+...+\frac{9}{97.100}\)
\(A=9\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{97.100}\right)\)
\(A=9.\frac{1}{3}\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...-\frac{1}{100}\right)\)
\(A=\frac{9}{3}\left(\frac{1}{1}-\frac{1}{100}\right)\)
\(A=3\left(\frac{99}{100}\right)=\frac{297}{100}\)
NP
1
K
27 tháng 6 2015
\(E=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{31.34}\)
\(E=\frac{2}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{31.34}\right)\)
\(E=\frac{2}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{31}-\frac{1}{34}\right)\)
\(E=\frac{2}{3}\left(1-\frac{1}{34}\right)\)
\(E=\frac{2}{3}.\frac{33}{34}\)
\(E=\frac{11}{17}\)
3 tháng 3 2023
=2/3(3/1*4+3/4*7+...+3/97*100)
=2/3(1-1/4+1/4-1/7+...+1/97-1/100)
=2/3*99/100
=198/300
=66/100
=33/50
