b: 6B=2*4*6+4*6*6+6*8*6+...+46*48*6+48*50*6

=2*4*6-2*4*6+4*6*8-4*6*8+...-44*46*48+46*48*50-46*48*50+48*50*52

=48*50*52

=>B=20800

d: 9D=1*4*9+4*7*9+...+46*49*9

=1*4*2+1*4*7-1*4*7+1*7*10-1*7*10+...+46*49*52-46*49*43

=1*2*4+46*49*52

=117216

=>D=13024

a: 

`#3107.101107`

1.

a)

`1/(1*4) + 1/(4*7) + 1/(7*10) + ... + 1/(100*103)`

`= 1/3 * (3/(1*4) + 3/(4*7) + 3/(7*10) + ... + 3/(100*103) )`

`= 1/3 * (1 - 1/4 + 1/4 - 1/7 + ... + 1/100 - 1/103)`

`= 1/3* (1 - 1/103)`

`= 1/3*102/103`

`= 34/103`

b)

`-1/3 + (-1/15) + (-1/35) + (-1/63) + ... + (-1/9999)`

`= - 1/3 - 1/15 - 1/35 - 1/63 - ... - 1/9999`

`= - (1/3 + 1/15 + 1/35 + ... + 1/9999)`

`= - (1/(1*3) + 1/(3*5) + 1/(5*7) + ... + 1/99*101)`

`= - 1/2 * (2/(1*3) + 2/(3*5) + 2/(5*7) + ... + 2/99*101)`

`= - 1/2* (1 - 1/3 + 1/3 - 1/5 + ... + 1/99 - 1/101)`

`= -1/2 * (1 - 1/101)`

`= -1/2*100/101`

`= -50/101`

2.

`3/(1*4) + 3/(4*7) + ... + 3/(94*97) + 3/(97*100)`

`= 1 - 1/4 + 1/4 - 1/7 + ... + 1/94 - 1/97 + 1/97 - 1/100`

`= 1-1/100`

`= 99/100`

\(A=1.4+4.7+7.10+....+97.100\)

\(2A=2.8+8.14+14.20+...+194.200\)

\(2A=2\left(1.4+4.7+7.10+...+97.100\right)\)

\(2A-A=2\left(1.4+4.7+...+97.100\right)-\left(1.4+4.7+...+97.100\right)\)

\(\Rightarrow A=2\)

Vậy.....

mk ko biết làm nhưng sai rồi bạn ơi

\(B=\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{2021.2014}\)

\(\Rightarrow B=\dfrac{1}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{2021}-\dfrac{1}{2014}\right)\)

\(\Rightarrow B=\dfrac{1}{3}.\left(1-\dfrac{1}{2014}\right)\)

\(\Rightarrow B=\dfrac{1}{3}.\dfrac{2013}{2014}=\dfrac{671}{2014}\)

\(B=\dfrac{1}{1\cdot4}+\dfrac{1}{4\cdot7}+...+\dfrac{1}{2021\cdot2024}\\ =\dfrac{1}{3}\cdot\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{2021\cdot2024}\right)\\ =\dfrac{1}{3}\cdot\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{2021}-\dfrac{1}{2024}\right)\\ =\dfrac{1}{3}\cdot\left(1-\dfrac{1}{2024}\right)\\ =\dfrac{1}{3}\cdot\dfrac{2023}{2024}\\ =\dfrac{2023}{6072}\)

\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x\left(x+3\right)}=\frac{125}{376}\)

=>\(3\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x\left(x+3\right)}\right)=3.\frac{125}{376}\)

=>\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{x\left(x+3\right)}=\frac{375}{376}\)

=>\(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{375}{376}\)

=>\(1-\frac{1}{x+3}=\frac{375}{376}\)

=>\(\frac{1}{x+3}=1-\frac{375}{376}\)

=>\(\frac{1}{x+3}=\frac{1}{376}\)

=>x+3=376

=>x=376-3

=>x=373

Vậy x=373

1/1+4 +1/4×7 +1/7×10+.....+1/x×(x+3)=16/49

\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{94.97}+\frac{3}{97.100}\)

\(=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{94}-\frac{1}{97}+\frac{1}{97}-\frac{1}{100}\)

\(=\frac{1}{1}-\frac{1}{100}=\frac{100}{100}-\frac{1}{100}=\frac{99}{100}\)

\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{94.97}+\frac{3}{97.100}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{94}-\frac{1}{97}+\frac{1}{97}-\frac{1}{100}\)

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}\)

Đặt biểu thức trên là A. Ta có:

3A = 3/1.4 + 3/4.7 + 3/7.10 + ... + 3/2016/2019

3A = 1-1/4 +1/4-1/7+1/7-1/10/+ ... + 1/2016-1/2019

3A = 1-1/2019=2018/2019

A =1009/2019

Ta có:

\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{2016.2019}\)

\(=\frac{1}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{2016.2019}\right)\)

\(=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+....+\frac{1}{2016}-\frac{1}{2019}\right)\)

\(=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{2019}\right)\)

\(=\frac{1}{3}.\frac{2018}{2019}\)

\(=\frac{2018}{6057}\)