Có thể đề sẽ là \(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+\dfrac{1}{10.13}+...+\dfrac{1}{97.100}\\ =\dfrac{1}{3}\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+\dfrac{3}{10.13}+...+\dfrac{3}{97.100}\right)\\ =\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\\ =\dfrac{1}{3}\left(1-\dfrac{1}{100}\right)=\dfrac{1}{3}.\dfrac{99}{100}=\dfrac{33}{100}\)

Bạn đánh lại chỗ 1/(j-4) nha.

KQ = 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=\(\frac{2}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+........+\frac{3}{97.100}\right)\)

=\(\frac{2}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...........+\frac{1}{97}-\frac{1}{100}\right)\)

=\(\frac{2}{3}.\left(1-\frac{1}{100}\right)\)

=\(\frac{2}{3}.\frac{99}{100}\)

=\(\frac{33}{50}\)

33/50

\(\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{97.100}\)

= \(\frac{2}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)

= \(\frac{2}{3}\left(1-\frac{1}{100}\right)\)

= \(\frac{2}{3}.\frac{99}{100}=\frac{33}{50}\)

\(D=\frac{2}{1.4}+\frac{2}{4.7}+...+\frac{2}{97.100}\)

\(3D=\frac{2.3}{1.4}+\frac{2.3}{4.7}+...+\frac{2.3}{97.100}\)

\(3D=2\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{97.100}\right)\)

\(3D=2\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\right)\)

\(3D=2\left(1-\frac{1}{100}\right)\)

\(3D=2\cdot\frac{99}{100}\)

\(3D=\frac{99}{50}\)

\(D=\frac{99}{50}:3\)

\(D=\frac{33}{50}\)