Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(=\dfrac{1}{8}\left(1+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}\right)\)
\(=\dfrac{1}{8}\cdot2\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}\right)\)
\(=\dfrac{1}{4}\cdot\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{6}-\dfrac{1}{7}\right)\)
\(=\dfrac{1}{4}\cdot\dfrac{6}{7}=\dfrac{3}{14}\)
=\(1+\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+...+\dfrac{1}{10\cdot12}\)
\(=1+\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{10}-\dfrac{1}{12}\right)\)
\(=1+\dfrac{1}{2}\cdot\dfrac{5}{12}=1+\dfrac{5}{24}=\dfrac{29}{24}\)
=1+12⋅4+14⋅6+...+110⋅121+12⋅4+14⋅6+...+110⋅12
=1+12(12−14+14−16+...+110−112)=1+12(12−14+14−16+...+110−112)
=1+12⋅512=1+524=2924
\(A=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{9900}\)
\(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\)
\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(A=1-\dfrac{1}{100}=\dfrac{99}{100}\)
\(B=\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+..+\dfrac{1}{195}\) ( là 195 ms đúng ! )
\(B=\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{13\cdot15}\)
\(B=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{13}-\dfrac{1}{15}\right)\)
\(B=\dfrac{1}{2}\left(1-\dfrac{1}{15}\right)=\dfrac{1}{2}\cdot\dfrac{14}{15}=\dfrac{7}{15}\)
\(C=\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+...+\dfrac{1}{98\cdot100}\)
Rồi làm tương tự cân b nha!
\(D=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{32}+\dfrac{1}{32}-\dfrac{1}{57}\)
\(+\dfrac{1}{57}-\dfrac{1}{87}\)
\(D=\dfrac{1}{3}-\dfrac{1}{87}=\dfrac{28}{87}\)
\(B=\left(1+\dfrac{1}{8}\right)\left(1+\dfrac{1}{15}\right)\left(1+\dfrac{1}{24}\right).....\left(1+\dfrac{1}{440}\right)\left(1+\dfrac{1}{483}\right)\)
\(B=\dfrac{9}{8}.\dfrac{16}{15}.\dfrac{25}{24}.....\dfrac{441}{440}.\dfrac{484}{483}\)
\(B=\dfrac{9.16.25.....441.484}{8.15.24.....440.483}\)
\(B=\dfrac{3.3.4.4.5.5.....21.21.22.22}{2.4.3.5.4.6.....20.22.21.23}\)
\(B=\dfrac{3.4.5.....21.22}{2.3.4.....20.21}.\dfrac{3.4.5.....21.22}{4.5.6.....22.23}\)
\(B=11.\dfrac{3}{23}=\dfrac{33}{23}\)
B = \(\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}.\dfrac{25}{24}...\dfrac{121}{120}.\dfrac{144}{143}\)
B = \(\dfrac{4.9.16.25...121.144}{3.8.15.24....120.143}\)
B = \(\dfrac{2.2.3.3.4.4.5.5...11.11.12.12}{1.3.2.4.3.5.4.6...10.12.11.13}\)
B = \(\dfrac{2.3.4.5...11.12}{1.2.3.4.5...10.11}.\dfrac{2.3.4.5...11.12}{3.4.5.6.7...12.13}\)
B = 12 . \(\dfrac{2}{13}\)
B = \(\dfrac{24}{13}\)
đây là tính nhanh à nếu tính bình thường thì tính may tính là ra
a) 17/23 . 8/16 . 23/17. (-80) . 3/4
= (17/23 . 23/17) . (8/16 . 3/4) . (-80)
= 1 . 3/8 . (-80)
= 3/8 . (-80)
= -30
b) 5/11 . 18/29 - 5/11 . 8/29 + 5/11 . 19/29
= 5/11 . (18/29 - 8/29 + 19/29)
= 5/11 . 1
= 5/11
c)(13/23 + 1313/2323 - 131313/232323).(1/3+1/4 -7/12)
= (13/23 + 1313/2323 - 131313/232323).0
= 0
d) 12/2x2 . 22/2x3 . 32/3x4 . 42/4x5 . 52/5x6 . 62/6x7 . 72/7x8 . 82/8x9 . 92/9x10
= 1/2 . 2/3 . 3/4 . 4/5 . 5/6 . 6/7 . 7/8 . 8/9 .9/10
= 1/10
Khó nhìn quá. Bạn thông cảm nhé! 
a) \(\dfrac{-3}{7}+\dfrac{15}{26}-\left(\dfrac{2}{13}-\dfrac{3}{7}\right)\\ =\dfrac{-3}{7}+\dfrac{15}{26}-\dfrac{2}{13}+\dfrac{3}{7}\\ =\left(\dfrac{-3}{7}+\dfrac{3}{7}\right)+\left(\dfrac{15}{26}-\dfrac{2}{13}\right)\\ =0+\left(\dfrac{15}{26}-\dfrac{4}{26}\right)\\ =0+\dfrac{11}{26}\\ =\dfrac{11}{26}\)
\(c)\dfrac{-11}{23}.\dfrac{6}{7}+\dfrac{8}{7}.\dfrac{-11}{23}-\dfrac{1}{23}\\=\dfrac{-1}{23}\left ( \dfrac{66}{7}+\dfrac{88}{7}+1 \right )\\ =\dfrac{-1}{23}.23=-1\)
Đặt \(A=\dfrac{1}{41}+\dfrac{1}{42}+\dfrac{1}{43}+\dfrac{1}{44}+...+\dfrac{1}{80}\)
\(=\left(\dfrac{1}{41}+\dfrac{1}{42}+...+\dfrac{1}{60}\right)+\) \(\left(\dfrac{1}{61}+\dfrac{1}{62}+...+\dfrac{1}{80}\right)\)
Nhận xét:
\(\dfrac{1}{41}+\dfrac{1}{42}+...+\dfrac{1}{60}>\dfrac{1}{60}+\dfrac{1}{60}+...+\dfrac{1}{60}\) \(=\dfrac{1}{3}\)
\(\dfrac{1}{61}+\dfrac{1}{62}+...+\dfrac{1}{80}>\dfrac{1}{80}+\dfrac{1}{80}+...+\dfrac{1}{80}\) \(=\dfrac{1}{4}\)
\(\Rightarrow A>\dfrac{1}{3}+\dfrac{1}{4}=\dfrac{7}{12}>\dfrac{1}{12}\)
Vậy \(\dfrac{1}{41}+\dfrac{1}{42}+\dfrac{1}{43}+...+\dfrac{1}{80}>\dfrac{1}{12}\) (Đpcm)
Ta có: \(A=1+\dfrac{1}{8}+\dfrac{1}{24}+\dfrac{1}{48}+\dfrac{1}{80}+\dfrac{1}{120}\)
\(\Leftrightarrow2A=2+\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+\dfrac{2}{8\cdot10}+\dfrac{2}{10\cdot12}\)
\(\Leftrightarrow2A=2+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{12}\)
\(\Leftrightarrow2A=2+\dfrac{1}{2}-\dfrac{1}{12}\)
\(\Leftrightarrow2A=\dfrac{24}{12}+\dfrac{6}{12}-\dfrac{1}{12}\)
\(\Leftrightarrow2A=\dfrac{29}{12}\)
hay \(A=\dfrac{29}{24}\)