ai tk mk thì mk tk lại

\(\dfrac{1}{4.7}+\dfrac{1}{7.10}+\dfrac{1}{10.13}+...+\dfrac{1}{37.40}< \dfrac{1}{5}\)

=\(\dfrac{3}{3}\left(\dfrac{1}{4.7}+\dfrac{1}{7.10}+\dfrac{1}{10.13}+...+\dfrac{1}{37.40}\right)\)

=\(\dfrac{1}{3}\left(\dfrac{3}{4.7}+\dfrac{3}{7.10}+\dfrac{3}{10.13}+...+\dfrac{3}{37.40}\right)\)

=\(\dfrac{1}{3}\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+...+\dfrac{1}{37}-\dfrac{1}{40}\right)\)

=\(\dfrac{1}{3}\left(\dfrac{1}{4}-\dfrac{1}{40}\right)\)

=\(\dfrac{3}{40}< \dfrac{1}{3}\)

1) 

A= \(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{39.40}\)

\(\Rightarrow A=\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-...+\frac{1}{39}-\frac{1}{40}\)

\(\Rightarrow A=\frac{1}{3}-\frac{1}{40}\)

=> A= 27/120

A = \(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{39.40}\)

= \(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{39}-\frac{1}{40}\)

= \(\frac{1}{3}-\frac{1}{40}\)

= \(\frac{37}{120}\)

B = \(\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{37.40}\)

= \(\frac{1}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{37}-\frac{1}{40}\right)\)

= \(\frac{1}{3}\left(\frac{1}{4}-\frac{1}{40}\right)\)

= \(\frac{1}{3}.\frac{9}{40}=\frac{3}{40}\)

C = \(\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{37.40}\)

= \(\frac{2}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{37}-\frac{1}{40}\right)\)

= \(\frac{2}{3}.\left(\frac{1}{4}-\frac{1}{40}\right)\)

= \(\frac{2}{3}.\frac{9}{40}=\frac{3}{20}\)

1) A = \(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{39.40}\)

A = \(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{39}-\frac{1}{40}\)

A = \(\frac{1}{3}-\frac{1}{40}\)

A = \(\frac{37}{120}\)

2) B = \(\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{7}\right)+\frac{1}{3}.\left(\frac{1}{7}-\frac{1}{10}\right)+...+\frac{1}{3}.\left(\frac{1}{37}-\frac{1}{40}\right)\)

B = \(\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{37}-\frac{1}{40}\right)\)

B = \(\frac{1}{3}.\frac{9}{40}\)

B = \(\frac{3}{40}\)

3) C = \(\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{37.40}\)

C = \(\frac{2}{3}.\left(\frac{1}{4}-\frac{1}{7}\right)+\frac{2}{3}.\left(\frac{1}{7}-\frac{1}{10}\right)+...+\frac{2}{3}.\left(\frac{1}{37}-\frac{1}{40}\right)\)

C = \(\frac{2}{3}.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{37}-\frac{1}{40}\right)\)

C = \(\frac{2}{3}.\left(\frac{1}{4}-\frac{1}{40}\right)\)

C = \(\frac{2}{3}.\frac{9}{40}\)

C = \(\frac{3}{20}\)

Ta có:

Đặt \(A=\)\(\dfrac{1}{4.7}+\dfrac{1}{7.10}+\dfrac{1}{10.13}+...+\dfrac{1}{604.607}< \dfrac{1}{2}\)

\(=\dfrac{1}{3}.\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+...+\dfrac{1}{604}-\dfrac{1}{607}\right)< \dfrac{1}{2}\)

\(=\dfrac{1}{3}.\left(\dfrac{1}{4}-\dfrac{1}{607}\right)< \dfrac{1}{2}\)

Vì \(\dfrac{1}{3}< \dfrac{1}{2}\) nên \(\dfrac{1}{3}.\left(\dfrac{1}{4}-\dfrac{1}{607}\right)< \dfrac{1}{2}\) 

Vậy \(A< \dfrac{1}{2}\)

............................... =) A < 1/2

\(A=\dfrac{1}{3}\left(\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{25\cdot28}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{25}-\dfrac{1}{28}\right)\)

\(=\dfrac{1}{3}\cdot\dfrac{6}{28}=\dfrac{2}{28}=\dfrac{1}{14}\)

`3A = 3/(4.7) + 3/(7.10) + .. + 3/(25.28)`

`3A = 1/4 - 1/7 + 1/7 - 1/10 +... + 1/25 - 1/28`

`3A = 3/14`

`A = 1/14.`

Nguyễn Huy Thắng giải sai rồi ,thế này mới đúng nè

1,\(\frac{1}{6}+\frac{1}{12}+.........+\frac{1}{72}\)

=\(\frac{1}{2.3}+\frac{1}{3.4}+.......+\frac{1}{8.9}\)

=\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.......+\frac{1}{8}-\frac{1}{9}\)

=\(\frac{1}{2}-\frac{1}{9}\)

=\(\frac{7}{18}\)

2,\(\frac{3}{1.4}+\frac{3}{4.7}+..........+\frac{3}{13.16}\)

=\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.........+\frac{1}{13}-\frac{1}{16}\)

=\(1-\frac{1}{16}\)

=\(\frac{15}{16}\)

2)đặt B= 3/1.4+3/4.7+3/7.10+3/10.13+3/13.16

\(B=3\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{16}\right)\)

\(B=3-\frac{15}{16}\)

\(B=\frac{45}{16}\)

1)

1/6+1/12+1/30+1/56+1/72

= 1/2x3+1/3x4+1/4x5+1/5x6+1/6x7+1/7x8

= 1/2-1/3+1/3-1/4+...+1/7-1/8

= 1/2-1/8

= 3/8

2)

3/1.4+3/4.7+3/7.10+3/10.13+3/13.16

= 1-1/4+1/4-1/7+..+1/13-1/16

= 1-1/16

= 15/16

mk nhé đầu tiên đó

\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+\dfrac{1}{10.13}+...+\dfrac{1}{x\left(x+3\right)}=\dfrac{34}{103}\)

\(\dfrac{1}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{34}{103}\)

\(\dfrac{1}{3}.\left(1-\dfrac{1}{x+3}\right)=\dfrac{34}{103}\)

\(1-\dfrac{1}{x+3}=\dfrac{34}{103}:\dfrac{1}{3}=\dfrac{34}{103}.3\)

\(1-\dfrac{1}{x+3}=\dfrac{102}{103}\)

\(\dfrac{1}{x+3}=1-\dfrac{102}{103}=\dfrac{103}{103}-\dfrac{102}{103}\)

\(\dfrac{1}{x+3}=\dfrac{1}{103}\)

\(\Rightarrow x+3=103\)

\(x=103-3\)

\(x=100\)

Vậy x = 100

=>[(1/4-1/7+1/7-1/10+...+1/37-1/40):3]-x=4/5

=>[(1/4-1/40):3]-x=4/5

=>(9/40:3)-x=4/5

=>3/40-x=4/5

=>x=3/40-4/5

=>x=-29/40

K nhe bn. Chinh xac roi day

A=\(\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{73.76}\)

3A=3(\(\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{73.76}\))

3A=\(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{73.76}\)

3A=\(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{73}+\frac{1}{76}\)

3A=\(\frac{1}{4}-\frac{1}{76}\)

3A=\(\frac{9}{38}\)

A=\(\frac{9}{38}\):3

A=\(\frac{3}{38}\)

đặt A=1/4.7+1/7.10+...+1/73.76

3A=1/4-1/7+1/7-1/10+...+1/ 73 -1/ 76
 

\(A=\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{73.76}\)

\(3A=\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{73.76}\)

\(3A=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{73}-\frac{1}{76}\)

\(3A=\frac{1}{4}-\frac{1}{76}\)

\(3A=\frac{9}{38}\)

\(A=\frac{3}{38}\)