\(\dfrac{1}{7}\left(\dfrac{7}{3.10}+\dfrac{7}{10.17}+...+\dfrac{7}{73.80}-\left(\dfrac{7}{2.9}+\dfrac{7}{9.16}+...+\dfrac{7}{23.30}\right)\right)\)

\(=\dfrac{1}{7}\left(\dfrac{1}{3}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{17}+...+\dfrac{1}{73}-\dfrac{1}{80}-\left(\dfrac{1}{2}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...+\dfrac{1}{23}-\dfrac{1}{30}\right)\right)\)

\(=\dfrac{1}{7}\left(\dfrac{1}{3}-\dfrac{1}{80}-\left(\dfrac{1}{2}-\dfrac{1}{30}\right)\right)\)

\(=\dfrac{1}{7}\left(\dfrac{77}{240}-\dfrac{7}{15}\right)=\dfrac{1}{7}.\left(-\dfrac{7}{48}\right)=-\dfrac{1}{48}\)

dỗiii c-các cậu vì ko lm hộ t-tớ -((

Đặt cái đó=A
Nhân 7A vào rồi áp dụng đến hết 7/n(n+7)=1/n-1/n+7, trùng cái nào thì khử

Ta có:

P=\(\frac{1}{3.10}\)+\(\frac{1}{10.17}\)+\(\frac{1}{17.24}\)+......+\(\frac{1}{73.80}\)-\(\frac{1}{2.9}\)-\(\frac{1}{9.16}\)-\(\frac{1}{16.23}\)-\(\frac{1}{23.30}\))

P=\(\frac{1}{7}\)\(\times\)(\(\frac{7}{3.10}\)+\(\frac{7}{10.17}\)+\(\frac{7}{17.24}\)+......\(\frac{7}{73.80}\)-\(\frac{7}{2.9}\)-\(\frac{7}{9.16}\)-\(\frac{7}{16.23}\)-\(\frac{7}{23.30}\))

P=\(\frac{1}{7}\)\(\times\)(\(\frac{1}{3}\)-\(\frac{1}{10}\)+\(\frac{1}{10}\)-\(\frac{1}{17}\)+.....+\(\frac{1}{73}\)-\(\frac{1}{80}\)-\(\frac{1}{2}\)-\(\frac{1}{9}\)-......-\(\frac{1}{23}\)-\(\frac{1}{30}\))

P=\(\frac{1}{7}\)\(\times\)(\(\frac{1}{3}\)-\(\frac{1}{80}\))-\(\frac{1}{7}\)(\(\frac{1}{2}\)-\(\frac{1}{30}\))

P=\(\frac{1}{7}\)\(\times\)(\(\frac{1}{3}\)-\(\frac{1}{80}\)-\(\frac{1}{2}\)+\(\frac{1}{30}\))

P=\(\frac{-7}{336}\)

Bài này mk ko tính máy tính nên ko chắc đâu

nhóm từng cái vào r` tính ra

Đặt biểu thức cần tính là A, ta có:

A=\(\dfrac{1}{7}\left(\dfrac{7}{3.10}+\dfrac{7}{10.17}+...+\dfrac{7}{73.80}\right)\)

Làm tg tự với những cái khác là ok

Nếu phân số thứ 2 là \(\frac{1}{10.17}\) thì làm như vậy nè

\(\frac{1}{3.10}+\frac{1}{10.17}+...+\frac{1}{73.80}-\frac{1}{2.9}-\frac{1}{9.16}-\frac{1}{16.23}-\frac{1}{23.30}\)

= \(\frac{1}{7}\left(\frac{1}{3}-\frac{1}{10}+\frac{1}{10}-\frac{1}{17}+...+\frac{1}{73}-\frac{1}{80}\right)-\left(\frac{1}{2.9}+\frac{1}{9.16}+\frac{1}{16.23}+\frac{1}{23.30}\right)\)

= \(\frac{1}{7}\left(\frac{1}{3}-\frac{1}{80}\right)-\frac{1}{7}\left(\frac{1}{2}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+\frac{1}{16}-\frac{1}{23}+\frac{1}{23}-\frac{1}{30}\right)\)

= \(\frac{1}{7}.\frac{77}{240}-\frac{1}{7}\left(\frac{1}{2}-\frac{1}{30}\right)=\frac{1}{7}.\frac{77}{240}-\frac{1}{7}.\frac{7}{15}\)

= \(\frac{11}{240}-\frac{1}{15}\)

= \(-\frac{1}{48}\)

Hồ Thu Giang nói đúng đấy

1/48

kick nha

có 100 lik-e tui cũng nói ko

\(A=\frac{1}{3\cdot10}+\frac{1}{10\cdot17}+\frac{1}{17\cdot24}+...+\frac{1}{73\cdot80}-\frac{1}{2\cdot9}-\frac{1}{9\cdot16}-\frac{1}{16\cdot23}-\frac{1}{23\cdot30}\)

\(A=\frac{1}{7}\left(\frac{7}{3\cdot10}+\frac{7}{10\cdot17}+\frac{7}{17\cdot24}+...+\frac{7}{73\cdot80}-\frac{7}{2\cdot9}-\frac{7}{9\cdot16}-\frac{7}{16\cdot23}-\frac{7}{23\cdot30}\right)\)

\(A=\frac{1}{7}\left(\frac{1}{3}-\frac{1}{10}+\frac{1}{10}-\frac{1}{17}+...+\frac{1}{73}-\frac{1}{80}-\frac{1}{2}+\frac{1}{9}-\frac{1}{9}+\frac{1}{16}-...-\frac{1}{23}-\frac{1}{30}\right)\)

\(A=\frac{1}{7}\left(\frac{1}{3}-\frac{1}{80}-\frac{1}{2}-\frac{1}{30}\right)\)

Đặt \(A=\frac{7^2}{2.9}+\frac{7^2}{9.16}+\frac{7^2}{16.23}+\frac{7^2}{23.30}\)

\(\Rightarrow A=7.\left(\frac{1}{2}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+\frac{1}{16}-\frac{1}{23}+\frac{1}{23}-\frac{1}{30}\right)\)

\(\Rightarrow A=7.\left(\frac{1}{2}-\frac{1}{30}\right)\)

\(\Rightarrow A=\frac{49}{15}\)

đặt biểu thức là B

Ta có công thức :

\(\frac{a}{b.c}=\frac{a}{c-b}.\left(\frac{1}{b}-\frac{1}{c}\right)\)

Dựa vào công thức, ta có :

\(B=7.\left(\frac{1}{2}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+.....+\frac{1}{23}-\frac{1}{30}\right)\)

\(B=7.\left(\frac{1}{2}-\frac{1}{30}\right)=7.\frac{7}{15}=\frac{49}{15}\)

Ai thấy đúng thì ủng hộ nha !!!

Đề là ntn:

\(A=49\left(\dfrac{1}{2.9}+\dfrac{1}{9.16}+\dfrac{1}{16.23}+...+\dfrac{1}{65.72}\right):\dfrac{1}{3}-\dfrac{7}{36}\)

\(A=7\left(\dfrac{1}{2}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{23}+...+\dfrac{1}{65}-\dfrac{1}{72}\right):\dfrac{1}{3}-\dfrac{7}{36}\)

\(A=7\left(\dfrac{1}{2}-\dfrac{1}{72}\right):\dfrac{1}{3}-\dfrac{7}{36}\)

\(A=7.\dfrac{35}{72}:\dfrac{1}{3}-\dfrac{7}{36}\)

\(A=\dfrac{245}{72}:\dfrac{1}{3}-\dfrac{7}{36}\)

\(A=\dfrac{735}{72}-\dfrac{7}{36}=\dfrac{735}{72}-\dfrac{14}{36}=\dfrac{721}{36}\)