Đáp án + Giải thích các bước giải:

7/1×5+7/5×9+7/9×13+7/13×17+7/17×21

=7/4×(4/1×5+4/5×9+4/9×13+4/13×17+4/17×21)

=7/4×(1−1/5+1/5−1/9+1/9−1/13+1/13−1/17+1/17−1/21)

=7/4×(1+0+0+0+0−1/21)

=7/4×(1−1/21)

=7/4×2021

=5/3

nhớ tick nha

\(\dfrac{7}{1\times5}+\dfrac{7}{5\times9}+\dfrac{7}{9\times13}+\dfrac{7}{13\times17}\)\(+\) \(\dfrac{7}{17\times21}\)

= \(\dfrac{7}{4}\times\)( \(\dfrac{4}{1\times5}\)\(+\) \(\dfrac{4}{5\times9}\)\(+\) \(\dfrac{4}{9\times13}\)\(+\)\(\dfrac{4}{13\times17}\)\(+\)\(\dfrac{4}{17\times21}\))

= 7 \(\times\)(\(\dfrac{1}{1}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{17}-\dfrac{1}{21}\))

= 7\(\times\)(\(\dfrac{1}{1}-\dfrac{1}{21}\))

= \(\dfrac{7}{4}\)\(\times\) \(\dfrac{20}{21}\)

= \(\dfrac{5}{3}\)

5/3

7/1x5+7/5x9+7/9x13+7/13x17+7/17x21=75,5 120 1609

ஜ۩۞۩ஜ Ủng hộ tui nhaaaa ஜ۩۞۩ஜ 

 Ai đi qua xin để lại 1 k !!!!!!

a) 1/1x5 + ... + 1/21x25

= 4 x (1-1/5 + 1/5 - 1/9 + ... + 1/21 - 1/25)

= 1/4 x (1 - 1/25)

= 1/4 x 24/25

= 6/25

\(A=\frac{7}{1\cdot5}+\frac{7}{5\cdot9}+...+\frac{7}{17\cdot21}=\)

\(\frac{4}{7}A=\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+...+\frac{4}{17\cdot21}=\)

\(\frac{4}{7}A=\left(1-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{9}\right)+...+\left(\frac{1}{17}-\frac{1}{21}\right)=\)

\(\frac{4}{7}A=1-\frac{1}{21}=\)

\(\frac{4}{7}A=\frac{20}{21}\)

\(A=\frac{20}{21}\div\frac{4}{7}\)

\(A=\frac{20}{21}\times\frac{7}{4}=\frac{140}{84}=\frac{5}{3}\)

\(\frac{7}{1.5}+\frac{7}{5.9}+\frac{7}{9.13}+...+\frac{7}{17.21}\)

\(=\frac{7}{4}.\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{17.21}\right)\)

\(=\frac{7}{4}.\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{17}-\frac{1}{21}\right)\)

\(=\frac{7}{4}.\left(1-\frac{1}{21}\right)\)

\(=\frac{7}{4}.\frac{20}{21}=\frac{5}{3}\)

 1/5 + 1/45 + 1/117 + 1/221 + 1/357+ .... + 1/159197

Đặt A=1/5 + 1/45 + 1/117 + 1/221 + 1/357+ .... + 1/159197

      A= 1/1x5 + 1/5x9 + 1/9x13 + 1/13x17 + .... + 1/397x401

   Ax4=4/1x5 + 4/5x9 + 4/9x13 + 4/13x17 + .... + 4/397x401

   Ax4=(1-1/5) + (1/5-1/9) + (1/9-1/13)+….+(1/397-1/401)

   Ax4=1-1/401

       A=100/401

Mà 1/4=100/400>100/401=A nên A<1/4.

gọi tổng này là một số A

ta có  

A=7/4x(1/1x5+1/5x9+1/9x13+1/13x17+1/17x21)

A= 7/4x(1-1/5+1/5-1/9+1/9-1/13+1/13-1/17+1/17-1/21)

A= 7/4x(1-1/21)=7/4x20/21

suy ra A=5/3

7/1×5 + 7/5x9 + 7/9x13 + 7/13x17 + 7/17x21

= 7/4x(4/1x5 + 4/5x9 + 4/9x13 + 4/13x17 + 4/17x21)

= 7/4x(1 - 1/5 + 1/5 - 1/9 + 1/9 - 1/13 + 1/13 - 1/17 + 1/17 - 1/21)

= 7/4x(1-1/20)

= 7/4x19/20

= 133/80

đặt A=7/1X5+7/5X9+7/9X13+7/13X17+7/17X21

Ta có:4A=4x(7/1x5+7/5x9+7/9x13+7/13x17+7/17x21)

4A=7x(4/1x5+4/5x9+4/9x13+4/13x17+4/17x21)

4A=7x(1-1/5+1/5-1/9+1/9-1/13+1/13-1/17+1/17-1/21)

4A=7x(1-1/21)

4A=7x20/21

A=20/3:4

A=20/3 x 1/4

A=5/3

Vậy A=5/3

Bài 2:

\(A=\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+\frac{1}{238}+\frac{1}{340}\)

\(3A=\frac{3}{2\times5}+\frac{3}{5\times8}+\frac{3}{8\times11}+\frac{3}{11\times14}+\frac{3}{14\times17}\)

\(3A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\)

\(3A=\frac{1}{2}-\frac{1}{17}=\frac{15}{34}\)

\(A=\frac{15}{34}\times\frac{1}{3}=\frac{5}{34}\)

Bài 2:

\(A=\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+\frac{1}{238}+\frac{1}{340}\)

\(3A=\frac{3}{2\times5}+\frac{3}{5\times8}+\frac{3}{8\times11}+\frac{3}{11\times14}+\frac{3}{14\times17}\)

\(A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\)

\(3A=\frac{1}{2}-\frac{1}{17}=\frac{15}{34}\)

\(A=\frac{15}{34}\times\frac{1}{3}=\frac{5}{34}\)

Cảm ơn cái đặc cầu

tính E phải ko bạn

E= 217/292

217/992 nha. Nếu đúng thì mình nha

Toán quá dễ. Tự túc là hạnh phúc mọi nhà bn nhé !

\(\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+\frac{3}{4.5}+\frac{3}{5.6}+...+\frac{3}{9.10}+\frac{77}{2.9}+\frac{77}{9.16}+\frac{77}{16.23}+...+\frac{77}{93.100}\)

Gọi \(\left(\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+......+\frac{3}{9.10}\right)\)là \(A\); \(\left(\frac{77}{2.9}+\frac{77}{9.16}+\frac{77}{16.23}+...+\frac{77}{93.100}\right)\)là B . Ta có : 

\(A=\frac{3}{1}.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\)

\(A=\frac{3}{1}.\left(\frac{1}{1}-\frac{1}{10}\right)\)

\(A=\frac{3}{1}\cdot\frac{9}{10}=\frac{27}{10}\)

\(B=\frac{77}{7}\left(\frac{1}{2}-\frac{1}{9}+\frac{1}{6}-\frac{1}{16}+\frac{1}{16}-\frac{1}{23}+....+\frac{1}{93}-\frac{1}{100}\right)\)

\(B=\frac{77}{7}\left(\frac{1}{2}-\frac{1}{100}\right)\)

\(B=\frac{77}{7}\cdot\frac{49}{100}=\frac{539}{100}\)

\(\Rightarrow\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+\frac{3}{4.5}+...+\frac{3}{9.10}+\frac{77}{2.9}+\frac{77}{9.16}+\frac{77}{16.23}+...+\frac{77}{93.100}=\frac{27}{10}+\frac{539}{100}=\frac{809}{100}\)

\(\frac{4}{3.6}+\frac{4}{6.9}+\frac{4}{9.12}+\frac{4}{12.15}\)

\(=\frac{4}{3}\cdot\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+\frac{1}{12}-\frac{1}{15}\right)\)

\(=\frac{4}{3}\cdot\left(\frac{1}{3}-\frac{1}{15}\right)\)

\(=\frac{4}{3}\cdot\frac{4}{15}=\frac{16}{45}\)

\(\frac{7}{1.5}+\frac{7}{5.9}+\frac{7}{9.13}+\frac{7}{13.17}+\frac{7}{17.21}\)

\(=\frac{7}{4}\cdot\left(\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+\frac{1}{17}-\frac{1}{21}\right)\)

\(=\frac{7}{4}\cdot\left(\frac{1}{1}-\frac{1}{21}\right)\)

\(=\frac{7}{4}\cdot\frac{20}{21}=\frac{5}{3}\)

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{110}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{10.11}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{10}-\frac{1}{11}\)

\(=\frac{1}{1}-\frac{1}{11}\)

\(=\frac{10}{11}\)

Sủa đề : \(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+\frac{1}{238}+\frac{1}{340}\)

\(=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+\frac{1}{17.20}\)

\(=\frac{1}{3}\cdot\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\right)\)

\(=\frac{1}{3}\cdot\left(\frac{1}{2}-\frac{1}{20}\right)\)

\(=\frac{1}{3}\cdot\frac{9}{20}=\frac{3}{20}\)