242 nha.

12+8drdhr547y

dap an la 242

A = 1(2+2)+2(3+2)+3(4+2)+.+99(100+2)

A = 1.2+1.2+2.3+2.2+3.4+3.2+.+99.100+99.2

A = (1.2+2.3+3.4+.+99.100)+2(1+2+3+.+99)

Phần còn lại bạn tự động não đi nha.

C= (4/1.3).(9/2.4).(16/3.5)...........(10000/99.101)
C=(4.9.16...........10000)/(1.3).(2.4)......(99.101)
C=(2^2.3^2.4^2..........100^2)/(1.2.3.4......99).(3.4.5.......101)
C=(2.3.4........100).(2.3.4......100)/(1.2.3.....99).(3.4.5....101)
Sau khi triệt tiêu ở tử và mẫu ta được:
C=(2.100)/101
C=200/101
 

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

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

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

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

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

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

\(=2\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{97.100}\right)\)

\(=\frac{2}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\right)\)

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

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

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

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

ủa ko phải câu đầu là như thế này hả bạn  \(A=\frac{2}{3}\left(\frac{2}{1.4}+\frac{2}{4.7}+...+\frac{2}{97.100}\right)\)

Đặt :

\(A=\dfrac{1}{1.4}+\dfrac{1}{4.7}+.................+\dfrac{1}{97.100}\)

\(3A=\dfrac{3}{1.4}+\dfrac{3}{4.7}+.................+\dfrac{3}{97.100}\)

\(3A=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+..........+\dfrac{1}{97}-\dfrac{1}{100}\)

\(3A=1-\dfrac{1}{100}\)

\(3A=\dfrac{99}{100}\)

\(\Rightarrow A=\dfrac{99}{100}:3\)

\(\Rightarrow A=\dfrac{33}{100}\)

~ Chúc bn học tốt ~

Ta có : \(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{97.100}\)

= 3( \(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{97.100}\) ).\(\dfrac{1}{3}\)

= (\(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{97.100}\)).\(\dfrac{1}{3}\)

= \(\left(\dfrac{4-1}{1.4}+\dfrac{7-4}{4.7}+\dfrac{10-7}{7.10}+...+\dfrac{100-97}{97.100}\right).\dfrac{1}{3}\) = \(\left(\dfrac{4}{1.4}-\dfrac{1}{1.4}+\dfrac{7}{4.7}-\dfrac{4}{4.7}+\dfrac{10}{7.10}-\dfrac{7}{7.10}+...+\dfrac{100}{97.100}-\dfrac{97}{97.100}\right).\dfrac{1}{3}\) = \(\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\right).\dfrac{1}{3}\) =\(\left(1-\dfrac{1}{100}\right).\dfrac{1}{3}\)

= \(\dfrac{99}{100}.\dfrac{1}{3}\)

= \(\dfrac{99}{100.3}\)

= \(\dfrac{33}{100}\)

Lời giải:

Đặt \(A=\dfrac{1}{1.4}+\dfrac{1}{4.7}+...+\dfrac{1}{97.100}\)

\(\Rightarrow A=\dfrac{1.3}{1.4.3}+\dfrac{1.3}{4.7.3}+...+\dfrac{1.3}{97.100.3}\)

\(\Rightarrow A=\dfrac{1}{3}.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{97.100}\right)\)

\(\Rightarrow A=\dfrac{1}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)

\(\Rightarrow A=\dfrac{1}{3}.\left(1-\dfrac{1}{100}\right)\)

\(\Rightarrow A=\dfrac{1}{3}.\dfrac{99}{100}\)

\(\Rightarrow A=\dfrac{33}{100}\)

\(C=1.4+2\left(4+1\right)+3\left(4+2\right)+...+99\left(4+98\right)\)

\(\Leftrightarrow C=1.4+2.4+1.2+3.4+2.3+...+99.4+98.99\)

\(C=\left(1.4+2.4+3.4+....+99.4\right)+\left(1.2+2.3+3.4+..98.99\right)\)

\(C=4\left(1+2+3+...+99\right)+\dfrac{1.2.3+2.3.3+3.4.3+...+98.99.3}{3}\)

\(C=\dfrac{4.\left(1+99\right).99}{2}+\dfrac{1.2.3+2.3\left(4-1\right)+...+98.99\left(100-97\right)}{3}\)

\(C=\dfrac{4.\left(1+99\right).99}{2}+\dfrac{1.2.3+2.3.4-1.2.3...+98.99.100-97.98.99}{3}\)

\(C=\dfrac{4.\left(1+99\right).99}{2}+\dfrac{98.99.100}{3}\)

\(C=19800+107800=127600\)

\(A=\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{x\left(x+3\right)}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+3}\)

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

\(=\frac{x+2}{x+3}=\frac{100}{101}\)

\(\Rightarrow x=98\)

\(A=\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{x.\left(x+3\right)}=\frac{100}{101}\)

\(A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{100}{101}\)

\(A=1-\frac{1}{x+3}=\frac{100}{101}\)

\(\frac{1}{x+3}=1-\frac{100}{101}=\frac{1}{101}\)

=> x + 3 = 101

=> x = 101 - 3

=> x = 98

Vậy x = 98

Ủng hộ mk nha ^_-

Baig iair 

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:

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

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

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

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

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

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

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

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

\(=\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{2016}-\frac{1}{2019}\right)\)

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

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

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

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

`S_1 = 5/(1.4) + 5/(4.7) +...+ 5/(97.100)`

`S_1 = 5 (1/(1.4) + 1/(4.7) +...+ 1/(97.100))`

`S_1 = 5/3 (3/(1.4) + 3/(4.7) +...+ 3/(97.100))`

`S_1 = 5/3 (1 - 1/4 + 1/4 - 1/7 + ...+ 1/97 - 1/100)`

`S_1 = 5/3 (1 - 1/100)`

`S_1 = 5/3 . 99/100`

`S_1 = 33/20`