1*4+4*7+7*10+...+97*100
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Có thể đề sẽ là \(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+\dfrac{1}{10.13}+...+\dfrac{1}{97.100}\\ =\dfrac{1}{3}\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+\dfrac{3}{10.13}+...+\dfrac{3}{97.100}\right)\\ =\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\\ =\dfrac{1}{3}\left(1-\dfrac{1}{100}\right)=\dfrac{1}{3}.\dfrac{99}{100}=\dfrac{33}{100}\)
\(\dfrac{2}{1\times4}+\dfrac{2}{4\times7}+\dfrac{2}{7\times10}+\cdot\cdot\cdot+\dfrac{2}{97\times100}\)
\(=\dfrac{2}{3}\times\left(\dfrac{3}{1\times4}+\dfrac{3}{4\times7}+\dfrac{3}{7\times10}+\cdot\cdot\cdot+\dfrac{3}{97\times100}\right)\)
\(=\dfrac{2}{3}\times\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\cdot\cdot\cdot+\dfrac{1}{97}-\dfrac{1}{100}\right)\)
\(=\dfrac{2}{3}\times\left(1-\dfrac{1}{100}\right)\)
\(=\dfrac{2}{3}\times\dfrac{99}{100}\)
\(=\dfrac{33}{50}\)
#\(Toru\)
=> 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
Ta đặt biểu thức là :
A = 2/1 x 4 + 2/4 x 7 + 2/7 x 10 + ... + 2/97 x 100
A = 2 - 2/4 + 2/4 - 2/7 + 2/7 - 2/10 + ... + 2/97 - 2/100
A = 2 - 2 /100
A = 99/50
A=2/3 x (1-1/4+1/4-1/7+......+1/97-1/100)
= 2/3 x (1-1/100)
= 2/3 x 99/100
= 33/50
A=\(\frac{2}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+........+\frac{3}{97.100}\right)\)
=\(\frac{2}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...........+\frac{1}{97}-\frac{1}{100}\right)\)
=\(\frac{2}{3}.\left(1-\frac{1}{100}\right)\)
=\(\frac{2}{3}.\frac{99}{100}\)
=\(\frac{33}{50}\)
\(\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{97.100}\)
= \(\frac{2}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)
= \(\frac{2}{3}\left(1-\frac{1}{100}\right)\)
= \(\frac{2}{3}.\frac{99}{100}=\frac{33}{50}\)
\(D=\frac{2}{1.4}+\frac{2}{4.7}+...+\frac{2}{97.100}\)
\(3D=\frac{2.3}{1.4}+\frac{2.3}{4.7}+...+\frac{2.3}{97.100}\)
\(3D=2\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{97.100}\right)\)
\(3D=2\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(3D=2\left(1-\frac{1}{100}\right)\)
\(3D=2\cdot\frac{99}{100}\)
\(3D=\frac{99}{50}\)
\(D=\frac{99}{50}:3\)
\(D=\frac{33}{50}\)
Đặt A = 1*4+4*7+7*10+...+97*100
=> 7A = 1*4*(7-0) + 7*10*(13 - 6)
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