Bài 1:

\(A=\dfrac{3}{1.4}+\dfrac{5}{4.9}+\dfrac{7}{9.16}+\dfrac{9}{16.25}+\dfrac{11}{25.36}\)

\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{25}+\dfrac{1}{25}-\dfrac{1}{36}\)

\(=1-\dfrac{1}{36}=\dfrac{35}{36}\)

\(B=\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{100.103}\)

\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\)

\(=1-\dfrac{1}{103}=\dfrac{102}{103}\)

\(C=\dfrac{3}{1.4}+\dfrac{6}{4.10}+\dfrac{9}{10.19}+\dfrac{12}{19.31}+\dfrac{15}{31.46}+\dfrac{18}{46.64}\)

\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{31}+\dfrac{1}{31}-\dfrac{1}{46}+\dfrac{1}{46}-\dfrac{1}{64}\)

\(=1-\dfrac{1}{64}=\dfrac{63}{64}\)

Bài 2: 

\(\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{49.50}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{49}-\dfrac{1}{50}\)

\(=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)

\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)

\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{25}\right)\)

\(=\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+...+\dfrac{1}{50}\left(đpcm\right)\)

Sửa đề: 1/1.4 (số hạng thứ 2) ➜ 1/4.7

Giải:

1/1.4+1/4.7+1/7.10+...+1/x.(x+3)=6/19

1/3.(3/1.4+3/4.7+3/7.10+...+3/x.(x+3))=6/19

1/3.(1/1-1/4+1/4-1/7+1/7-1/10+...+1/x-1/x+3)=6/19

1/3.(1/1-1/x+3)                            =6/19

       1/1-1/x+3                             =6/19:1/3

       1/1-1/x+3                             =18/19

              1/x+3                            =1/1-18/19

               1/x+3                           =1/19

⇒x+3=19

      x=19-3

      x=16

Chúc bạn học tốt!

`#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`

\(\dfrac{3}{2}A=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{94.97}\)

\(\dfrac{3}{2}A=\dfrac{4-1}{1.4}+\dfrac{7-4}{4.7}+\dfrac{10-7}{7.10}+...+\dfrac{97-94}{94.97}\)

\(\dfrac{3}{2}A=\dfrac{4}{1.4}-\dfrac{1}{1.4}+\dfrac{7}{4.7}-\dfrac{4}{4.7}+\dfrac{10}{7.10}-\dfrac{7}{7.10}+...+\dfrac{97}{94.97}-\dfrac{94}{94.97}\)

\(\dfrac{3}{2}A=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{94}-\dfrac{1}{97}\)

\(\dfrac{3}{2}A=1-\dfrac{1}{97}=\dfrac{96}{97}\)

⇒ A = \(\dfrac{96}{97}:\dfrac{3}{2}=\dfrac{64}{97}\)

Câu B cách làm tương tự, thắc mắc gì bạn cứ hỏi nhé.

Ta có :

Đặt

(1)

Từ (1), suy ra:

Vậy

~ Học tốt ~

Ta có :

Đặt

(1)

Từ (1), suy ra:

Vậy

~ Học tốt ~

\(\Leftrightarrow2^{6x-1-2}=2^9\)

=>6x-3=9

=>6x=12

=>x=2

`@` `\text {Answer}`

`\downarrow`

`h)`

\(2^{6x-1}\div4=8^3\)

`=>`\(2^{6x-1}=\left(2^3\right)^3\cdot4\)

`=>`\(2^{6x-1}=2^9\cdot2^2\)

`=>`\(2^{6x-1}=2^{11}\)

`=> 6x - 1 = 11`

`=> 6x = 11 +1`

`=> 6x = 12`

`=> x = 12 \div 6`

`=> x = 2`

Vậy, `x = 2.`

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