Tính tổng B = \(\frac{1}{2.9}+\frac{1}{9.16}+\frac{1}{16.23}+...+\frac{1}{100.107}+\frac{1}{107.114}\)
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Nếu phân số thứ 2 là \(\frac{1}{10.17}\) thì làm như vậy nè
\(\frac{1}{3.10}+\frac{1}{10.17}+...+\frac{1}{73.80}-\frac{1}{2.9}-\frac{1}{9.16}-\frac{1}{16.23}-\frac{1}{23.30}\)
= \(\frac{1}{7}\left(\frac{1}{3}-\frac{1}{10}+\frac{1}{10}-\frac{1}{17}+...+\frac{1}{73}-\frac{1}{80}\right)-\left(\frac{1}{2.9}+\frac{1}{9.16}+\frac{1}{16.23}+\frac{1}{23.30}\right)\)
= \(\frac{1}{7}\left(\frac{1}{3}-\frac{1}{80}\right)-\frac{1}{7}\left(\frac{1}{2}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+\frac{1}{16}-\frac{1}{23}+\frac{1}{23}-\frac{1}{30}\right)\)
= \(\frac{1}{7}.\frac{77}{240}-\frac{1}{7}\left(\frac{1}{2}-\frac{1}{30}\right)=\frac{1}{7}.\frac{77}{240}-\frac{1}{7}.\frac{7}{15}\)
= \(\frac{11}{240}-\frac{1}{15}\)
= \(-\frac{1}{48}\)
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
taị mk ko tính máy tính lên sai.
bn thông cảm nha. thường ngày hay dùng máy tính quá nên tính sai thì bn thông cảm
A = 7 (7 / 2.9 + 7 / 9.16 + .......... + 7/65.72)
A=7( 1/2 - 1/9 +1/9 - 1/16 +......+1/65 - 1/72)
A= 7 ( 1/2 -1/72)
A= 7 . 35/72
A=245/72
\(A=\frac{7^2}{2.9}+\frac{7^2}{9.16}+\frac{7}{16.23}+.....+\frac{7^2}{65.72}\)
=\(7.\left(\frac{1}{2}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+\frac{1}{16}-\frac{1}{23}+....+\frac{1}{65}-\frac{1}{72}\right)\)
=\(7.\left(\frac{1}{2}-\frac{1}{72}\right)\)
=\(7.\frac{35}{72}\)
=\(\frac{245}{72}\)
\(C=\frac{7^2}{2.9}+\frac{7^2}{9.16}+\frac{7^2}{16.23}+...+\frac{7^2}{65.72}\)
\(C=\frac{7^2}{7}.\left(\frac{1}{2}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+\frac{1}{16}-\frac{1}{23}+...+\frac{1}{65}-\frac{1}{72}\right)\)
\(C=7.\left(\frac{1}{2}-\frac{1}{72}\right)\)
\(C=7.\frac{35}{72}=\frac{245}{72}\)
Ta có : \(C=\frac{7^2}{2.9}+\frac{7^2}{9.16}+\frac{7^2}{16.23}+.....+\frac{7^2}{65.72}\)
\(\Rightarrow C=7\left(\frac{7}{2.9}+\frac{7}{9.16}+\frac{7}{16.23}+.....+\frac{7}{65.72}\right)\)
\(\Rightarrow C=7\left(\frac{1}{2}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+.....+\frac{1}{65}-\frac{1}{72}\right)\)
\(\Rightarrow C=7\left(\frac{1}{2}-\frac{1}{72}\right)\)
\(\Rightarrow C=7.\frac{35}{72}=\frac{245}{72}\)
\(C=\frac{7^2}{2.9}+\frac{7^2}{9.16}+\frac{7^2}{16.23}+....+\frac{7^2}{65.72}\)
\(C=\frac{7^2}{7}\cdot\left(\frac{1}{2}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+\frac{1}{16}-\frac{1}{23}+....+\frac{1}{65}-\frac{1}{72}\right)\)
\(C=7\cdot\left(\frac{1}{2}-\frac{1}{72}\right)\)
\(C=7\cdot\frac{35}{72}=\frac{245}{72}\)
C = 49(1/2.9 ... 1/65.72)
C = 49(1/2 - 1/9 +....+ 1/65 - 1/72)
C = 49( 1/2 - 1/72)
C = bạn tự tính nhé
Có j không hiểu thì Ib mình
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 !!!
Ta có:
C = \(\frac{7^2}{2.9}+\frac{7^2}{9.16}+\frac{7^2}{16.23}+...+\frac{7^2}{65.72}\)
=> C = \(7.\left(\frac{7}{2.9}+\frac{7}{9.16}+\frac{7}{16.23}+...+\frac{7}{65.72}\right)\)
=> C = \(7.\left(\frac{1}{2}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+\frac{1}{16}-\frac{1}{23}+...+\frac{1}{65}-\frac{1}{72}\right)\)
=> C = \(7.\left(\frac{1}{2}-\frac{1}{72}\right)\)
=> C = \(7.\frac{35}{72}=\frac{245}{72}\)
Nhìn kĩ là ra thôi :
\(\frac{7^2}{2.9}+\frac{7^2}{9.16}+...+\frac{7^2}{65.72}\)
= \(7\left(\frac{7}{2.9}+\frac{7}{9.16}+...+\frac{7}{65.72}\right)\)
= \(7\left(\frac{1}{2}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+...+\frac{1}{65}-\frac{1}{72}\right)\)
= \(7\left(\frac{1}{2}-\frac{1}{72}\right)\)
= \(7.\frac{35}{72}=3\frac{29}{72}\)
\(A=\frac{1}{3\cdot10}+\frac{1}{10\cdot17}+\frac{1}{17\cdot24}+...+\frac{1}{73\cdot80}-\frac{1}{2\cdot9}-\frac{1}{9\cdot16}-\frac{1}{16\cdot23}-\frac{1}{23\cdot30}\)
\(A=\frac{1}{7}\left(\frac{7}{3\cdot10}+\frac{7}{10\cdot17}+\frac{7}{17\cdot24}+...+\frac{7}{73\cdot80}-\frac{7}{2\cdot9}-\frac{7}{9\cdot16}-\frac{7}{16\cdot23}-\frac{7}{23\cdot30}\right)\)
\(A=\frac{1}{7}\left(\frac{1}{3}-\frac{1}{10}+\frac{1}{10}-\frac{1}{17}+...+\frac{1}{73}-\frac{1}{80}-\frac{1}{2}+\frac{1}{9}-\frac{1}{9}+\frac{1}{16}-...-\frac{1}{23}-\frac{1}{30}\right)\)
\(A=\frac{1}{7}\left(\frac{1}{3}-\frac{1}{80}-\frac{1}{2}-\frac{1}{30}\right)\)
Hướng dẫn:
\(7B=\dfrac{9-2}{2.9}+\dfrac{16-9}{9.16}+\dfrac{23-16}{16.23}+...+\dfrac{107-100}{100.107}+\dfrac{114-107}{107.114}\)
\(=\dfrac{1}{2}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{23}+...+\dfrac{1}{100}-\dfrac{1}{107}+\dfrac{1}{107}-\dfrac{1}{114}\)
\(=\dfrac{1}{2}-\dfrac{1}{114}=...\)
\(B=\frac{1}{2.9}+\frac{1}{9.16}+\frac{1}{16.23}+...+\frac{1}{100.107}+\frac{1}{107.114}\)
\(B=\frac{1}{7}\left(\frac{1}{2}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+\frac{1}{16}-\frac{1}{23}+...+\frac{1}{100}-\frac{1}{107}+\frac{1}{107}-\frac{1}{114}\right)\)
\(B=\frac{1}{7}\left(\frac{1}{2}-\frac{1}{114}\right)\)
\(B=\frac{1}{7}\left(\frac{57}{114}-\frac{1}{114}\right)\)
\(B=\frac{1}{7}.\frac{56}{114}\)
\(B=\frac{1}{7}.\frac{28}{57}\)
\(B=\frac{4}{57}\)