1: \(=\dfrac{1}{29\cdot30}-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{28\cdot29}\right)\)

\(=\dfrac{1}{29\cdot30}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{28}-\dfrac{1}{29}\right)\)

\(=\dfrac{1}{29\cdot30}-\dfrac{28}{29}=\dfrac{1-28\cdot30}{870}=\dfrac{-859}{870}\)

a, 68 + 42 x 5 - 625 : 25

  = 68 + 210    -  25

=    278  - 25

=     253

b, 15 x { 37 + 8 + 20 } + 15 x { 13 + 22 }

  = 15 x   65               + 15 x    35

  = 15 x { 65 + 35 }

  = 15 x  100

  =  1500

c, 125 x 56 x 3

   = 7000  x 3 

   =    21000

d, 249 - { 49 - 100 }

  = 249 - 49 + 100

  = 200  + 100

  =  300

e, 1/2 + 1/4 + 1/ 8 + 1/32 + 1/64 + 1/128

   = 1/1 - 1/2 + 1/2 - 1/4 + 1/4 - 1/8 + 1/8 - 1/32 + 1/32 - 1/64 + 1/64 - 1/128

   = 1/1 - 1/128

   =   127/128

A, \(68+42\times5-625:25\)

= 68 +210 - 25

= 278 - 25

= 253

B,\(15\times\left(37+8+20\right)+15\times\left(13+22\right)\)

= \(15\times\left(37+8+20+13+22\right)\)

=\(15\times100\)

= 1500

C,\(125\times56\times3\)

= \(125\times8\times7\times3\)

=  \(1000\times21\)

= 21000

D, 249 - { 49 - 100 }

= 249 - 49 -100

= 200 - 100

= 100

E, \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)(mình sửa lại đề chút xíu nha, nó cứ bị sai sai sao ấy)

= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-\frac{1}{32}+\frac{1}{32}-\frac{1}{64}+\frac{1}{64}-\frac{1}{128}\) 

= \(1-\frac{1}{128}\)

=\(\frac{128}{128}-\frac{1}{128}\)

= \(\frac{127}{128}\)

A = 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56

A = 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + 1/6.7 + 1/7.8

A = 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 + 1/7 - 1/8

A = 1 + ( -1/2 + 1/2 ) + ( -1/3 + 1/3 ) + ( -1/4 + 1/4 ) + ( -1/5 + 1/5 ) + ( -1/6 + 1/6 ) + ( -1/7 + 1/7 ) - 1/8

A = 1 + 0 + 0 + 0 + 0 + 0 + 0 - 1/8

A = 1 - 1/8

A = 7/8

* Sửa đề tí nhé 

B = 3/2 - 5/6 + 7/12 - 9/20 + 11/30 - 13/42 + 15/56

B = 3/1.2 - 5/2.3 + 7/3.4 - 9/4.5 + 11/5.6 - 13/6.7 + 15/7.8

B = 3 - 3/2 - 5/2 - ( -5/3 ) + 7/3 - 7/4 - 9/4 - ( -9/5 ) + 11/5 - 11/6 - 13/6 - ( -13/7 ) + 15/7 - 15/8

B = 3 - 3/2 - 5/2 + 5/3 + 7/3 - 7/4 - 9/4 + 9/5 + 11/5 - 11/6 - 13/6 + 13/7 + 15/7 - 15/8

B = 3 + ( -3/2 - 5/2 ) + ( 5/3 + 7/3 ) + ( -7/4 - 9/4 ) + ( 9/5 + 11/5 ) + ( -11/6 - 13/6 ) + ( 13/7 + 15/7 ) - 15/8

B = 3 + -4 + 4 + -4 + 4 + -4 + 4 - 15/8

B = 3 + 0 + 0 + 0 - 15/8

B = 3 - 15/8

B = 9/8

em lớp 6  nha

B= 1/2 + 1/6 + 1/12 +1/20 + 1/30 + 1/42 + 1/56 + 1/72

B= 1/1*2 + 1/2*3 + 1/3*4 + 1/4*5 + 1/5*6 + 1/6*7 + 1/7*8 + 1/8*9

B=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9

B=1+0-0-0-0-0-0-0-1/9

B=1-1/9

B=8/9

k em nha

phần A em chịu

còn câu dưới đó e

a: \(A=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^7\)

=>\(2\cdot A=1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^6\)

=>\(2A-A=1-\left(\dfrac{1}{2}\right)^7=1-\dfrac{1}{128}=\dfrac{127}{128}\)

=>\(A=\dfrac{127}{128}\)

b: \(B=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{10\cdot11}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{10}-\dfrac{1}{11}\)

\(=1-\dfrac{1}{11}=\dfrac{10}{11}\)

1/100 - 1/90 - 1/72 - 1/56 - 1/42 - 1/30 - 1/20 - 1/12 - 1/6 - 1/2 = -0,833

\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)

\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)

\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)

\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)

\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)

\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)

\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)

A= 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56 + 1/72 + 1/90 

=1/(1.2)+1/(2.3)+1/(3.4)+1/(4.5) +1/(5.6)+1/(6.7)+1/(7.8) +1/(8.9)+1/(9.10) 

=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5.+1/5-1/6... +1/9-1/10 

=1-1/10 

=9/10

Me too ! 

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}=x-\left(2+4+..+100\right)\)

Gọi \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\)là \(A\)

         \(\left(2+4+...+100\right)\)là \(B\). Ta có :  

\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\)

\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{9}-\frac{1}{10}\)

\(A=\frac{1}{1}-\frac{1}{10}=\frac{9}{10}\)

Số số hạng của \(B\)là: \(\left(100-2\right)\div2+1=50\)

Tổng của \(B\)là: \(\left(2+100\right)\times50\div2=2550\)

\(\Rightarrow\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}=x-\left(2+4+..+100\right)=\frac{9}{10}=x-2550\)

\(\Rightarrow x=2550+\frac{9}{10}=2550+0,9=2550,9\)