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

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

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

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

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

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

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

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

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

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

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

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

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

 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.

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

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Đặt \(A=\frac{7}{1\cdot5}+\frac{7}{5\cdot9}+\frac{7}{9\cdot13}+\frac{7}{13\cdot17}+\frac{7}{17\cdot21}\)

\(\frac{4}{7}A=\frac{4}{7}\left(\frac{7}{1\cdot5}+\frac{7}{5\cdot9}+\frac{7}{9\cdot13}+\frac{7}{13\cdot17}+\frac{7}{17\cdot21}\right)\)

\(\frac{4}{7}A=\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+\frac{4}{17\cdot21}\)

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

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

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

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

Đặt biểu thức trên là A

\(\frac{4xA}{7}=\frac{4}{1x5}+\frac{4}{5x9}+\frac{4}{9x13}+\frac{4}{13x17}+\frac{4}{17x21}\)

\(\frac{4xA}{7}=\frac{5-1}{1x5}+\frac{9-5}{5x9}+\frac{13-9}{9x13}+\frac{17-13}{13x17}+\frac{21-17}{17x21}\)

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

\(\frac{4xA}{7}=1-\frac{1}{21}=\frac{20}{21}\Rightarrow A=\frac{20}{21}.\frac{7}{4}=\frac{5}{3}\)

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

\(E=\dfrac{3}{1x5}+\dfrac{3}{5x9}+...+\dfrac{3}{121x125}\)

\(\dfrac{4}{3}xE=1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{121}-\dfrac{1}{125}\)

\(\dfrac{4}{3}xE=1-\dfrac{1}{125}\)

\(E=\dfrac{124}{125}x\dfrac{3}{4}=\dfrac{93}{125}\)

`1/5 . 4/7 + 3/7 . 1/5 -1/5`

`=1/5 . 4/7 + 3/7 . 1/5 -1/5 . 1`

`=1/5 . ( 4/7+3/7-1)`

`=1/5 . ( 7/7-1)`

`= 1/5 . 0`

`=0`

\(\dfrac{1}{5}\times\dfrac{4}{7}+\dfrac{3}{7}\times\dfrac{1}{5}-\dfrac{1}{5}=\dfrac{1}{5}\times\left(\dfrac{4}{7}+\dfrac{3}{7}-1\right)=\dfrac{1}{5}\times0=0\)

= \(\dfrac{1}{5}\).(\(\dfrac{4}{7}\)+\(\dfrac{3}{7}\)+1)

= \(\dfrac{1}{5}\)+ 0

\(\dfrac{1}{5}\)