Trần Thùy Dung nó đã bảo \(990\ne99\cdot100\) rùi mà vẫn tách như v

=\(\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)

=\(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)

=\(\frac{1}{4}-\frac{1}{100}\)

=\(\frac{24}{100}=\frac{6}{25}\)

Bạn ơi sai rồi \(990\ne99.100\)

\(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+...+\frac{1}{9900}\)

\(=\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+...+\frac{1}{99\cdot100}\)

\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{99}-\frac{1}{100}\)

\(=\frac{1}{4}-\frac{1}{100}\)

\(=\frac{24}{100}=\frac{6}{25}\)

Đặt A=1/20+1/30+1/42+1/56+...+1/930

 =>A=1/4.5+1/5.6+1/6.7+...+1/30.31

=>A=1/4-1/5+1/5-1/6+...+1/30-1/31

=>A=1/4-1/31

=>A=31/124-4/124

=>A=27/124

Vây A=27/124

đề bài sai à bn phải là 1/930 chứ

(1/2+1/4+1/8+1/16):x=1/2+1/6+1/12+.....+1/132

15/16 : x = 1/1x2+1/2x3+1/3x4+.........+1/11x12

15/16 : x = 1-1/2+1/2-1/3+1/3-1/4+........+1/11-1/12

15/16 :x = 1-1/12

15/16 : x = 11/12

           x = 15/16 : 11/12

           x= 45/44

linhchi buithi, bạn ơi số 990 hình như thiếu một số 0 thì phải hay sao ý. Mình cứ thấy thiếu cái gì đó.

\(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{990}\)

\(=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}\)

\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{100}\)

\(=\frac{1}{4}-\frac{1}{100}\)

\(=\frac{25}{100}-\frac{1}{100}=\frac{6}{25}\)

\(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+....+\frac{1}{990}\)

=>\(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}\)

=>\(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)

=>\(\frac{1}{4}-\frac{1}{100}\)

=>\(\frac{6}{25}\)

`@` `\text {Ans}`

`\downarrow`

`a)`

`13/50 + 9% + 41/100 + 0,24`

`= 0,26 + 0,09 + 0,41 + 0,24`

`= (0,26 + 0,24) + (0,09 + 0,41)`

`= 0,5 + 0,5`

`= 1`

`b)`

`2018 \times 2020 - 1/2017 + 2018 \times 2019`

`= 2018 \times (2020 + 2019) - 1/2017`

`= 2018 \times 4039 - 1/2017`

`= 8150702`

`c)`

`1/2 + 1/6 + 1/12 + 1/20 +1/30 +1/42`

`=`\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+\dfrac{1}{6\times7}\)

`=`\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{6}-\dfrac{1}{7}\)

`=`\(1-\dfrac{1}{7}\)

`= 6/7`

\(a,\dfrac{13}{50}+9\%+\dfrac{41}{100}+0,24\\ 0,26+0,09+0,41+0,24\\ =\left(0,26+0,24\right)+\left(0,09+0,41\right)\\ =0,5+0,5\\ =1\\ b,2018\times2020-\dfrac{1}{2017}+2018\times2019\\ =2018\times\left(2020+2019\right)-\dfrac{1}{2017}\\ =2018\times4039-\dfrac{1}{2017}\\ =3150702-\dfrac{1}{2017}\\ c,\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}\\ =1-\dfrac{1}{2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}.........+\dfrac{1}{6}-\dfrac{1}{7}\\ =1-\dfrac{1}{7}\\ =\dfrac{6}{7}\)

Cảm ơn các bạn đã trả lời giúp mình câu hỏi nha

\(A=\dfrac{2}{1x3}+\dfrac{2}{3x5}+\dfrac{2}{5x7}+...+\dfrac{2}{21x23}\)

\(A=2x\left(\dfrac{1}{1x3}+\dfrac{1}{3x5}+\dfrac{1}{5x7}+...+\dfrac{1}{21x23}\right)\)

\(A=2x\dfrac{1}{2}x\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{21}-\dfrac{1}{23}\right)\)

\(A=1-\dfrac{1}{23}\)

\(A=\dfrac{22}{23}\)

\(B=\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\)

\(B=\dfrac{1}{2x3}+\dfrac{1}{3x4}+\dfrac{1}{4x5}+\dfrac{1}{5x6}+\dfrac{1}{6x7}+\dfrac{1}{7x8}+\dfrac{1}{8x9}+\dfrac{1}{9x10}\)

\(B=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)

\(B=\dfrac{1}{2}-\dfrac{1}{10}\)

\(B=\dfrac{5}{10}-\dfrac{1}{10}\)

\(B=\dfrac{4}{10}\)

\(B=\dfrac{2}{5}\)

\(C=\dfrac{1}{28}+\dfrac{1}{70}+\dfrac{1}{130}+\dfrac{1}{208}\)

\(C=\dfrac{1}{4x7}+\dfrac{1}{7x10}+\dfrac{1}{10x13}+\dfrac{1}{13x16}\)

\(C=\dfrac{1}{3}x\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{16}\right)\)

\(C=\dfrac{1}{3}x\left(\dfrac{1}{4}-\dfrac{1}{16}\right)\)

\(C=\dfrac{1}{3}x\left(\dfrac{4}{16}-\dfrac{1}{16}\right)\)

\(C=\dfrac{1}{3}x\dfrac{3}{16}\)

\(C=\dfrac{1}{16}\)

Ta có :

\(\frac{1}{12}=\frac{1}{12}\)

\(\frac{1}{13}< \frac{1}{12}\)

\(\frac{1}{14}< \frac{1}{12}\)

\(........\)

\(\frac{1}{17}< \frac{1}{12}\)

Cộng vế với vế ta có :

\(\frac{1}{12}+\frac{1}{13}+....+\frac{1}{17}< \frac{1}{12}+\frac{1}{12}+...+\frac{1}{12}\)(có 6 số \(\frac{1}{12}\))\(=\frac{6}{12}=\frac{1}{2}\)

Vậy \(\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+....+\frac{1}{17}< \frac{1}{2}\)

\(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+...+\dfrac{1}{50}+\dfrac{1}{990}\)

\(=\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{50}+\dfrac{1}{990}???\)

Quy luật của vế sau "..." sai, bạn xem lại đề bài!

Nếu đúng đề thì sẽ như sau:

\(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+...+\dfrac{1}{9900}\)

Đề bài đúng là như vậy.

Giải:

\(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+...+\dfrac{1}{9900}\)

\(=\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+...+\dfrac{1}{99.100}\)

\(=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=\dfrac{1}{4}-\dfrac{1}{100}\)

\(=\dfrac{25-1}{100}\)

\(=\dfrac{24}{100}\)

\(=\dfrac{6}{25}\)