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Đặt \(A=\frac{5}{4.6}+\frac{5}{6.8}+.....+\frac{5}{48.50}\)
\(\Leftrightarrow\frac{5}{2}A=\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}+.....+\frac{2}{48.50}\)
\(\Leftrightarrow\frac{5}{2}A=\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+....+\frac{1}{48}-\frac{1}{50}\)
\(\Leftrightarrow\frac{5}{2}A=\frac{1}{4}-\frac{1}{50}\)
\(\Leftrightarrow\frac{5}{2}A=\frac{23}{100}\Rightarrow A=\frac{23}{100}.\frac{2}{5}=\frac{23}{250}\)
S = 2*4+4*6+6*8+...+46*48+48*50
S6 = 2*4*6+4*6*6+6*8*6+........................+46*48*6+48*50*6
S6=2*4*(6-0)+4*6*(8-2)+6*8*(10-4)+.................................+46*48*(50-44)+48*50*(52-46)
S6 = 2*4*6+4*6*8-2*4*6+6*8*10-4*6*8+..........................................+46*48*50-44*46*48+48*50*52-46*48*50
S6 = 48*50*52=124800
S=124800/6=20800
\(S=2\cdot4+4\cdot6+...+48\cdot50\)
\(S=2\left(1\cdot2+2\cdot3+...+24\cdot25\right)\)
\(\Rightarrow3S=2\left(1\cdot2\left(3-0\right)+2\cdot3\left(4-1\right)+...+24\cdot25\left(26-23\right)\right)\)
\(\Rightarrow3S=2\left(1\cdot2\cdot3-0\cdot1\cdot2+2\cdot3\cdot4-1\cdot2\cdot3+...+24\cdot25\cdot26-23\cdot24\cdot25\right)\)
\(\Rightarrow3S=2\cdot24\cdot25\cdot26\)
\(\Rightarrow S=2\cdot8\cdot25\cdot26=10400\)
\(\Rightarrow6S=10400\cdot6=62400\)
\(\frac{5}{2\cdot4}+\frac{5}{4\cdot6}+\frac{5}{6\cdot8}+.....+\frac{5}{48\cdot60}\)
\(=\frac{5}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+.....+\frac{1}{48}-\frac{1}{50}\right)\)
\(=\frac{5}{2}\left(\frac{1}{2}-\frac{1}{50}\right)\)
Tự tính nốt :p
Ta có:
`7/(-5) = (-7)/5 < -1`
`-1 < (-2)/5 < 0`
`0 < (-4)/(-5) = 4/5`
Thứ tự giảm dần là: `4/5; (-2)/5; 7/(-5)`
a) `1/3 - 1/4 : 2/5 = 1/3 - 5/8 = -7/24`
b) `6/7-(5/6+1/3)-(2/3+1/7) = 6/7-5/6-1/3-2/3-1/7`
`=(6/7-1/7)-(1/3+2/3)-5/6`
`=5/7-1-5/6`
`=-47/42`
c) `-5/9 . 2/5 + 4 5/9 + 5/9 . (-3/5)`
`= -5/9 . 2/5 + 4 + 5/9 + (-5/9) . 3/5`
`=-5/9 . (2/5 + 3/5-1) + 4`
`=-5/9 . 0 +4`
`=4`
d) 3 1/2 - (5 4/7 - 1 1/2) : 0,75`
`=7/2 - (39/7 - 3/2) : 3/4`
`= 7/2 - 57/14 : 3/4`
`=7/2 - 38/7`
`=-27/14`
\(a.\dfrac{2}{3}+\dfrac{4}{3}:\dfrac{-2}{3}=\dfrac{2}{3}+\left(-2\right)=\dfrac{-4}{3}\)
\(b.3\dfrac{4}{5}-\left(2\dfrac{1}{4}+1\dfrac{4}{5}\right)\\ =3\dfrac{4}{5}-2\dfrac{1}{4}-1\dfrac{4}{5}\\ =\left(3\dfrac{4}{5}-1\dfrac{4}{5}\right)-2\dfrac{1}{4}\\ =2-2\dfrac{1}{4}=\dfrac{1}{4}\)
\(c.\dfrac{-3}{5}.\dfrac{4}{7}+\dfrac{3}{7}.\dfrac{-3}{5}+\dfrac{3}{5}\\ =\dfrac{-3}{5}\left(\dfrac{4}{7}+\dfrac{3}{7}\right)+\dfrac{3}{5}\\ =\dfrac{-3}{5}+\dfrac{3}{5}=0\)
a) \(\dfrac{2}{5}+\dfrac{4}{3}:\dfrac{-2}{3}\)
\(=\dfrac{2}{5}+\dfrac{4}{3}.\dfrac{-3}{2}\)
\(=\dfrac{2}{5}+-2\)
\(=\dfrac{2}{5}+\dfrac{-10}{5}\)
\(=\dfrac{-8}{5}\)
b: \(=\dfrac{2}{5}+\dfrac{3}{5}:\dfrac{9-10}{15}-\dfrac{7}{2}\)
\(=\dfrac{4-35}{10}+\dfrac{3}{5}\cdot\dfrac{15}{-1}\)
\(=\dfrac{-31}{10}-9=\dfrac{-31}{10}-\dfrac{90}{10}=-\dfrac{121}{10}\)
c: \(=\dfrac{48-5}{12}\cdot\dfrac{1}{3}+\dfrac{7}{36}=\dfrac{43}{36}+\dfrac{7}{36}=\dfrac{50}{36}=\dfrac{25}{18}\)
d: \(=\dfrac{17}{6}:\dfrac{6}{5}+\dfrac{-7}{12}\)
\(=\dfrac{85}{36}-\dfrac{7}{12}=\dfrac{85}{36}-\dfrac{21}{36}=\dfrac{64}{36}=\dfrac{16}{9}\)
b: =12+5/14-3-5/7-5-5/14
=4-5/7
=28/7-5/7=23/7
c: =(-2/5-11/10)+(7/11-7/11)
=-4/10-11/10=-15/10=-3/2
\(a,\dfrac{5}{9}\cdot\dfrac{7}{13}+\dfrac{5}{9}\cdot\dfrac{8}{13}-\dfrac{5}{13}\cdot\dfrac{2}{9}\)
\(=\dfrac{5}{9}\cdot\dfrac{7}{13}+\dfrac{5}{9}\cdot\dfrac{8}{13}-\dfrac{2}{13}\cdot\dfrac{5}{9}\)
\(=\dfrac{5}{9}\cdot\left(\dfrac{7}{13}+\dfrac{8}{13}-\dfrac{2}{13}\right)\)
\(=\dfrac{5}{9}\cdot\dfrac{14}{13}\)
\(=\dfrac{70}{117}\)
\(d,\dfrac{1}{2}+\dfrac{-2}{3}+\dfrac{1}{6}+\dfrac{-2}{5}\)
\(=\left(\dfrac{1}{2}+\dfrac{-2}{3}+\dfrac{1}{6}\right)+\dfrac{-2}{5}\)
\(=0+\dfrac{-2}{5}\)
\(=\dfrac{-2}{5}\)
a: =27/45-20/45=7/45
b: \(=\dfrac{3}{5}+\dfrac{30}{40}=\dfrac{3}{5}+\dfrac{3}{4}=\dfrac{12}{20}+\dfrac{15}{20}=\dfrac{27}{20}\)
c: \(=\dfrac{8}{13}\left(\dfrac{7}{2}-\dfrac{5}{2}+1\right)=\dfrac{8}{13}\cdot2=\dfrac{16}{13}\)
d: \(=\dfrac{9}{23}\left(\dfrac{5}{17}-\dfrac{22}{17}\right)+11+\dfrac{9}{23}=11\)
a) \(\dfrac{3}{5}+\dfrac{-4}{9}=\dfrac{27}{45}+\dfrac{-20}{45}=\dfrac{7}{45}\)
b) \(\dfrac{3}{5}+\dfrac{2}{5}.\dfrac{15}{8}=1.\dfrac{15}{8}=\dfrac{15}{8}\)
c) \(\dfrac{7}{2}.\dfrac{8}{13}+\dfrac{8}{13}.\dfrac{-5}{2}+\dfrac{8}{13}=\dfrac{8}{13}.\left(\dfrac{7}{2}+\dfrac{-5}{2}\right)=\dfrac{8}{13}.1=\dfrac{8}{13}\)
d) \(\dfrac{-5}{17}.\dfrac{-9}{23}+\dfrac{9}{23}.\dfrac{-22}{17}+11\dfrac{9}{23}=\dfrac{9}{23}.\left(\dfrac{-5}{17}+\dfrac{-22}{17}\right)=\dfrac{-243}{391}\)
\(C=\dfrac{5}{2.4}+\dfrac{5}{4.6}+\dfrac{5}{6.8}+...+\dfrac{5}{48.50}\)
Đặt \(C=\dfrac{5}{2}.\left(\dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+...+\dfrac{1}{48.50}\right)\)
\(C=\dfrac{5}{2}.\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{48}-\dfrac{1}{50}\right)\)
\(C=\dfrac{5}{2}.\left(\dfrac{1}{2}-\dfrac{1}{50}\right)\)
\(C=\dfrac{5}{2}.\left(\dfrac{25}{50}-\dfrac{1}{50}\right)\)
\(C=\dfrac{5}{2}.\dfrac{24}{50}\)
\(C=\dfrac{12}{10}=\dfrac{6}{5}\)
Vậy \(C=\dfrac{6}{5}\)