Tính giá trị của biểu thức:
P=\(\frac{1}{3.10}+\frac{1}{10.17}+\frac{1}{17.24}+...+\frac{1}{73.80}-\frac{1}{2.9}-\frac{1}{9.16}-\frac{1}{16.23}-\frac{1}{23.30}\)
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Nếu phân số thứ 2 là \(\frac{1}{10.17}\) thì làm như vậy nè
\(\frac{1}{3.10}+\frac{1}{10.17}+...+\frac{1}{73.80}-\frac{1}{2.9}-\frac{1}{9.16}-\frac{1}{16.23}-\frac{1}{23.30}\)
= \(\frac{1}{7}\left(\frac{1}{3}-\frac{1}{10}+\frac{1}{10}-\frac{1}{17}+...+\frac{1}{73}-\frac{1}{80}\right)-\left(\frac{1}{2.9}+\frac{1}{9.16}+\frac{1}{16.23}+\frac{1}{23.30}\right)\)
= \(\frac{1}{7}\left(\frac{1}{3}-\frac{1}{80}\right)-\frac{1}{7}\left(\frac{1}{2}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+\frac{1}{16}-\frac{1}{23}+\frac{1}{23}-\frac{1}{30}\right)\)
= \(\frac{1}{7}.\frac{77}{240}-\frac{1}{7}\left(\frac{1}{2}-\frac{1}{30}\right)=\frac{1}{7}.\frac{77}{240}-\frac{1}{7}.\frac{7}{15}\)
= \(\frac{11}{240}-\frac{1}{15}\)
= \(-\frac{1}{48}\)
1/3.10+1/10.17+......+1/73.80 - 1/2.9 - 1/9.16 - 1/16.23 - 1/23.30
= (7/3.10+7/10.17+......+7/73.80) : 7 - (7/2.9 + 7/9.16 + 7/16.23 + 7/23.30) : 7
= (1/3-1/10+1/10-1/17+...+1/73-1/80) : 7 - (1/2-1/9+1/9-1/16+1/16-1/23+1/23-1/30) : 7
=(1/3-1/80) : 7 - (1/2-1/30) : 7
= 77/240 : 7 - 7/15 : 7
=11/240 - 1/15
= -1/48
Nhấn đúng cho mk nha!!!!!!!!!!!!!
\(A=\frac{1}{3\cdot10}+\frac{1}{10\cdot17}+\frac{1}{17\cdot24}+...+\frac{1}{73\cdot80}-\frac{1}{2\cdot9}-\frac{1}{9\cdot16}-\frac{1}{16\cdot23}-\frac{1}{23\cdot30}\)
\(A=\frac{1}{7}\left(\frac{7}{3\cdot10}+\frac{7}{10\cdot17}+\frac{7}{17\cdot24}+...+\frac{7}{73\cdot80}-\frac{7}{2\cdot9}-\frac{7}{9\cdot16}-\frac{7}{16\cdot23}-\frac{7}{23\cdot30}\right)\)
\(A=\frac{1}{7}\left(\frac{1}{3}-\frac{1}{10}+\frac{1}{10}-\frac{1}{17}+...+\frac{1}{73}-\frac{1}{80}-\frac{1}{2}+\frac{1}{9}-\frac{1}{9}+\frac{1}{16}-...-\frac{1}{23}-\frac{1}{30}\right)\)
\(A=\frac{1}{7}\left(\frac{1}{3}-\frac{1}{80}-\frac{1}{2}-\frac{1}{30}\right)\)
Đặt 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
\(\dfrac{1}{7}\left(\dfrac{7}{3.10}+\dfrac{7}{10.17}+...+\dfrac{7}{73.80}-\left(\dfrac{7}{2.9}+\dfrac{7}{9.16}+...+\dfrac{7}{23.30}\right)\right)\)
\(=\dfrac{1}{7}\left(\dfrac{1}{3}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{17}+...+\dfrac{1}{73}-\dfrac{1}{80}-\left(\dfrac{1}{2}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...+\dfrac{1}{23}-\dfrac{1}{30}\right)\right)\)
\(=\dfrac{1}{7}\left(\dfrac{1}{3}-\dfrac{1}{80}-\left(\dfrac{1}{2}-\dfrac{1}{30}\right)\right)\)
\(=\dfrac{1}{7}\left(\dfrac{77}{240}-\dfrac{7}{15}\right)=\dfrac{1}{7}.\left(-\dfrac{7}{48}\right)=-\dfrac{1}{48}\)
\(C=\frac{7^2}{2.9}+\frac{7^2}{9.16}+\frac{7^2}{16.23}+...+\frac{7^2}{65.72}\)
\(C=\frac{7^2}{7}.\left(\frac{1}{2}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+\frac{1}{16}-\frac{1}{23}+...+\frac{1}{65}-\frac{1}{72}\right)\)
\(C=7.\left(\frac{1}{2}-\frac{1}{72}\right)\)
\(C=7.\frac{35}{72}=\frac{245}{72}\)
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:
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