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PH
1
GK
30 tháng 3 2018
\(\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+...+\frac{1}{9700}=\frac{0,33x}{2009}\)
\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{97.100}=\frac{0,33x}{2009}\)
\(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}=\frac{0,33x}{2009}\)
\(\frac{1}{1}-\frac{1}{100}=\frac{0,33x}{2009}\)
\(\frac{100}{100}-\frac{1}{100}=\frac{0,33x}{2009}\)
\(\frac{99}{100}=\frac{0,33x}{2009}\)
\(\Rightarrow2009.99=100.0,33x\)
\(\Rightarrow2009.99=33x\)
\(\Rightarrow2009.99:33=x\)
\(\Rightarrow2009.3=x\)
\(\Rightarrow6027=x\)
Vậy \(x=6027\)(MK KO CHẮC NÓ ĐÚNG NHÉ )
MA
31 tháng 8 2016
\(A=\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+\frac{1}{130}+...+\frac{1}{9700}\)
\(A=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{97.100}\)
\(A=\frac{3}{3}\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{97.100}\right)\)
\(A=\frac{1}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{97.100}\right)\)
\(A=\frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(A=\frac{1}{3}\left(1-\frac{1}{100}\right)\)
\(A=\frac{1}{3}.\frac{99}{100}=\frac{33}{100}\)
GK
27 tháng 3 2018
\(\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+\frac{1}{130}+...+\frac{1}{9700}\)
\(=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{97.100}\)
\(=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\)
\(=\frac{1}{1}-\frac{1}{100}\)
\(=\frac{100}{100}-\frac{1}{100}\)
\(=\frac{99}{100}\)
NT
1
XO
7 tháng 6 2020
Ta có : \(\frac{1}{4}+\frac{1}{28}+....+\frac{1}{9700}=\frac{0,33x}{2009}\)
=> \(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{97.100}=\frac{0.99x}{2009}\)
=> \(\frac{1}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{97.100}\right)=\frac{0,33x}{2009}\)
=> \(\frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\right)=\frac{0,33x}{2009}\)
=> \(\frac{1}{3}\left(1-\frac{1}{100}\right)=\frac{0,33x}{2009}\)
=> \(\frac{33}{100}=\frac{0,33x}{2009}\Rightarrow33.2009=100.0,33x\)
=> 33.2009 = 33x
=> x = 2009
7 tháng 6 2020
Thanks bn nhìu nha, mình sẽ K cho bn ngay. Bn kb với mình nha.
T
14 tháng 4 2019
c) \(A=\frac{6}{4}+\frac{6}{28}+\frac{6}{70}+\frac{6}{130}+\frac{6}{208}\)
\(=\frac{6}{1.4}+\frac{6}{4.7}+\frac{6}{7.10}+\frac{6}{10.13}+\frac{6}{13.16}\)
\(=2\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}\right)\)
\(=2\left(1-\frac{1}{16}\right)\)
\(=2.\frac{15}{16}\)
\(=\frac{15}{8}\)
Vậy A=\(\frac{15}{8}\)
KN
14 tháng 4 2019
a) \(\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+...+\frac{3^2}{97.100}\)
\(=3\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\right)\)
\(=3\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(=3\left(1-\frac{1}{100}\right)\)
\(=3.\frac{99}{100}=\frac{297}{100}\)
LK
1
KC
29 tháng 6 2018
mk lỡ lm lộn bài của bn huỳnh kim đạt ở bài dưới nha
mk xin lỗi !
10 tháng 5 2017
\(2A=2\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)
\(2A=1+\frac{1}{2}+...+\frac{1}{2^9}\)
\(2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)
\(A=1-\frac{1}{2^{10}}=\frac{2^{10}-1}{2^{10}}=\frac{1023}{1024}\)
BẤM ĐÚNG NHÉ
10 tháng 5 2016
\(a,\frac{-7}{25}.\frac{11}{13}+\frac{-7}{25}.\frac{2}{13}-\frac{18}{25}\)
\(=\frac{-7}{25}.\left(\frac{11}{13}+\frac{2}{13}\right)-\frac{18}{25}=\frac{-7}{25}-\frac{18}{25}=-1\)
\(b,\frac{5}{7}.\frac{1}{3}-\frac{5}{7}.\frac{1}{4}-\frac{5}{7}.\frac{1}{12}=\frac{5}{7}.\left(\frac{1}{3}-\frac{1}{4}-\frac{1}{12}\right)=\frac{5}{7}.\left(\frac{4}{12}-\frac{3}{12}-\frac{1}{12}\right)\)
\(=\frac{5}{7}.0=0\)
10 tháng 5 2016
c)\(5\frac{2}{5}.4\frac{2}{7}+5\frac{5}{7}.5\frac{2}{5}=\frac{27}{5}.\frac{30}{7}+\frac{40}{7}.\frac{27}{5}=\frac{27}{5}.\left(\frac{30}{7}+\frac{40}{7}\right)\)
\(=\frac{27}{5}.10=27.2=54\)
\(d,75\%-1\frac{1}{2}+0,5:\frac{5}{12}-\left(\frac{-1}{2}\right)^2=\frac{3}{4}-\frac{3}{2}+\frac{1}{2}.\frac{12}{5}-\frac{1}{4}\)
\(=\left(\frac{3}{4}-\frac{1}{4}\right)-\frac{3}{2}+\frac{6}{5}=\frac{1}{2}-\frac{3}{2}+\frac{6}{5}=-1+\frac{6}{5}=\frac{-5}{5}+\frac{6}{5}=\frac{1}{5}\)
A = 1/4 + 1/28 + 1/70 +...+ 1/9700
A = 1/1.4 + 1/4.7 + 1/7.10 +...+ 1/97.100
3A = 3/1.4 + 3/4.7 + 3/7.10 +...+ 3/97.100
3A = 1 - 1/100
3A = 99/100
A=99/100:3=33/100
\(=\frac{1}{1.4}+\frac{1}{4.7}+..+\frac{1}{97.100}\)
\(=\frac{1}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{97.100}\right)\)
\(=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{100}\right)\)
\(=\frac{1}{3}.\frac{99}{100}=\frac{33}{100}\)