Ta có :

1/1.4+1/4.7+...+1/91.94

=1/3.(1/1-1/4+...+1/91-1/94)

=1/3.(1/1-1/94)

=1/3.93/94

=31/94

Ta có :

1/1.4+1/4.7+...+1/91.94

=1/3.(1/1-1/4+...+1/91-1/94)

=1/3.(1/1-1/94)

=1/3.93/94

=31/94

Ta có :

1/1.4+1/4.7+...+1/91.94

=1/3.(1/1-1/4+...+1/91-1/94)

=1/3.(1/1-1/94)

=1/3.93/94

=31/94

1/1.4+1/4.7+1/7.10+...+1/91.94

=1/3.(3/1.4+3/4.7+3/7.10+...+3/91.94)

=1/3.(1-1/4+1/4-1/7+1/7-1/10+...+1/91-1/94)

=1/3.(1-1-94)

=1/3.(93/94)

=31/94

Đặt A= 1/1*4+1/4*7+1/7*10+....+1/91*94

3A= 3/1*4+3/4*7+3/7*10+....+3/91*94

3A=1/1-1/4+1/4-1/7+1/7-1/10+............+1/91-1/94

3A=1-1/94=93/94=>A=93/94*1/3=31/94

=31/94 k mình nha bạn 

1/1x4+1/4x7+1/7x10+..+1/91x97

=1-1/4+1/4-1/7+1/7-1/10+...-1/91+1/97

=1-1/97

=96/97

\(A=\dfrac{1}{1\times4}+\dfrac{1}{4\times7}+...+\dfrac{1}{37\times40}\\ =\dfrac{1}{3}\times\left(\dfrac{3}{1\times4}+\dfrac{3}{4\times7}+...+\dfrac{3}{37\times40}\right)\\ =\dfrac{1}{3}\times\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{37}-\dfrac{1}{40}\right)\\ =\dfrac{1}{3}\times\left(1-\dfrac{1}{40}\right)\\ =\dfrac{1}{3}\times\dfrac{39}{40}\\ =\dfrac{13}{40}\)

1/1*4 + 1/4*7 + 1/7*10 + ... + 1/97*100

= 1/3(3/1*4 + 3/4*7 + 3/7*10 + ... + 3/97*100)

= 1/3(1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + .... + 1/97 - 1/100)

= 1/3(1 - 1/100)

= 1/3*99/100

= 33/100

trả lời 

=33/100

chúc bn

học tốt

D = \(\dfrac{1}{1.4}\) + \(\dfrac{1}{4.7}\) + \(\dfrac{1}{7.10}\)+...+ \(\dfrac{1}{91.94}\)

D = \(\dfrac{1}{1}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) -  \(\dfrac{1}{7}\) +  \(\dfrac{1}{7}\) - \(\dfrac{1}{10}\)+...+ \(\dfrac{1}{91}\) - \(\dfrac{1}{94}\)

D = \(\dfrac{1}{1}\) - \(\dfrac{1}{94}\)

D = \(\dfrac{93}{94}\)

\(S=\frac{1}{1\times4}+\frac{1}{4\times7}+\frac{1}{7\times10}+...+\frac{1}{94\times97}+\frac{1}{97\times100}\)

\(S=\frac{1}{3}\times\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{94}-\frac{1}{97}+\frac{1}{97}-\frac{1}{100}\right)\)

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

\(S=\frac{1}{3}\times\frac{99}{100}\)

\(S=\frac{33}{100}\)

1/1.4+1/4.7+1/7.10+1/10.13+1/13.16

=1/3.(3/1.4+3/4.7+3/7.10+3/10.13+3/13.16)

=1/3.(1/1-1/4+1/4-1/7+1/7-1/10+1/10-1/13+1/13-1/16)

=1/3.(1/1-1/16)

=1/3.(16/16-1/16)=1/3.15/16=5/16

\(=\frac{1}{3}x\left(\frac{3}{1x4}+\frac{3}{4x7}+...+\frac{3}{77x80}\right)\)

\(=\frac{1}{3}x\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{77}-\frac{1}{80}\right)\)

\(=\frac{1}{3}x\left(\frac{1}{1}-\frac{1}{80}\right)\)

\(=\frac{1}{3}\times\frac{79.}{80}\)

\(=\frac{79}{240}\)

Tk giúp mk nha cảm ơn !!

\(=\frac{79}{240}\)

=1−14 +14 −110 +...+119 −122 

=1−122 

 =  \(\frac{21}{22}\)

k cho mình nha chắc chắn đúng 100 %

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{10}+...+\frac{1}{19}-\frac{1}{22}\)

\(=1-\frac{1}{22}\)

\(=\frac{21}{22}\)

\(=\frac{1}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{16.19}+\frac{3}{19.22}\right)\)

\(=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{20}\right)\)

\(=\frac{1}{3}\left(1-\frac{1}{20}\right)\)

\(=\frac{1}{3}.\frac{19}{20}\)

\(=\frac{19}{60}\)