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9 tháng 1 2016

0,84 tich nha

9 tháng 1 2016

0,6 x 1,4 = 0,84

         tick nha VŨ THỊ HUYỀN TRANG

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!

7 tháng 5 2021

thank

3 tháng 9 2015

a, \(\left(\frac{5}{8}.0,6-5:3\frac{1}{3}\right).\left(\frac{1}{5}-1,4\right).\left(-5\right)^2.x=120\)

\(\left(\frac{5}{8}.\frac{3}{5}-5:\frac{10}{3}\right)\left(\frac{1}{5}-\frac{7}{5}\right).25x=120\)

\(\left(\frac{3}{8}-\frac{3}{2}\right)\left(-\frac{6}{5}\right).25x=120\)

\(\left(-\frac{9}{8}\right).\left(-\frac{6}{5}\right).25x=120\)

\(\frac{27}{20}.25x=120\)\(25x=120:\frac{27}{20}=\frac{800}{9}\)

\(x=\frac{800}{9}:25=\frac{32}{9}\)

đợi tí đi nấu cơm đã 

20 tháng 8 2023

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

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

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

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

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

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

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

18 tháng 8 2023

\(A=\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{61\cdot64}+\dfrac{3}{64\cdot67}\)

\(A=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{61}-\dfrac{1}{64}+\dfrac{1}{64}-\dfrac{1}{67}\)

\(A=1-\dfrac{1}{67}\) < 1

=> A<1

18 tháng 8 2023

Ta có:

\(A=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{61.64}+\dfrac{3}{64.67}\)

\(=3.\dfrac{1}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{61}-\dfrac{1}{64}+\dfrac{1}{64}-\dfrac{1}{67}\right)\)

\(=3.\left(1-\dfrac{1}{67}\right)\)

\(=3.\dfrac{66}{67}\)

\(=\dfrac{198}{67}\)

Vì \(\dfrac{198}{67}\) có tử lớn hơn mẫu nên \(\dfrac{198}{67}>1\)

Vậy \(A>1\)

18 tháng 9 2021

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: 

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

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

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

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

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

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

 

18 tháng 9 2021

thanks bạn nhayeu

2 tháng 10 2023

`#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`

15 tháng 4 2017

mình ko biết k nha mình đang âm