1/4+1/28+/1/70+...+1/9700=0,33x/2009
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\(\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+...+\frac{1}{9700}=\frac{0,33x}{2009}\)
\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{97.100}=\frac{0,33x}{2009}\)
\(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}=\frac{0,33x}{2009}\)
\(\frac{1}{1}-\frac{1}{100}=\frac{0,33x}{2009}\)
\(\frac{100}{100}-\frac{1}{100}=\frac{0,33x}{2009}\)
\(\frac{99}{100}=\frac{0,33x}{2009}\)
\(\Rightarrow2009.99=100.0,33x\)
\(\Rightarrow2009.99=33x\)
\(\Rightarrow2009.99:33=x\)
\(\Rightarrow2009.3=x\)
\(\Rightarrow6027=x\)
Vậy \(x=6027\)(MK KO CHẮC NÓ ĐÚNG NHÉ )
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
Thanks bn nhìu nha, mình sẽ K cho bn ngay. Bn kb với mình nha.
\(3M=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\)
\(3M=\frac{4-1}{1.4}+\frac{7-4}{4.7}+...+\frac{100-97}{97.100}\)
\(3M=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\)
\(3M=1-\frac{1}{100}\)
\(3M=\frac{99}{100}\)
\(M=\frac{33}{100}\)
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
\(=\frac{1}{1.4}+\frac{1}{4.7}+..+\frac{1}{97.100}\)
\(=\frac{1}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{97.100}\right)\)
\(=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{100}\right)\)
\(=\frac{1}{3}.\frac{99}{100}=\frac{33}{100}\)
\(A=\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+\frac{1}{130}+...+\frac{1}{9700}\)
\(A=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{97.100}\)
\(A=\frac{3}{3}\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{97.100}\right)\)
\(A=\frac{1}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{97.100}\right)\)
\(A=\frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(A=\frac{1}{3}\left(1-\frac{1}{100}\right)\)
\(A=\frac{1}{3}.\frac{99}{100}=\frac{33}{100}\)
\(\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+\frac{1}{130}+...+\frac{1}{9700}\)
\(=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{97.100}\)
\(=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\)
\(=\frac{1}{1}-\frac{1}{100}\)
\(=\frac{100}{100}-\frac{1}{100}\)
\(=\frac{99}{100}\)
\(\dfrac{3}{2}xA=\dfrac{3}{1x4}+\dfrac{3}{4x7}+\dfrac{3}{7x10}+...+\dfrac{3}{97x100}=\)
\(=\dfrac{4-1}{1x4}+\dfrac{7-4}{4x7}+\dfrac{10-7}{7x10}+...+\dfrac{100-97}{97x100}=\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}=\)
\(=1-\dfrac{1}{100}=\dfrac{99}{100}\Rightarrow A=\dfrac{99}{100}x\dfrac{2}{3}=\dfrac{33}{50}\)
=> 3/1.4+3/4.7+.....+3/97.100 = 0,99x/2014
=> 1-1/4+1/4-1/7+....+1/97-1/100 = 0,99x/2014
=> 0,99x/2014 = 1-1/100 = 99/100
=> x = 99/100 : 0,99/2014 = 2014
Vậy x = 2014
Tk mk nha