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\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\)
\(=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=1-\dfrac{1}{10}\)
\(=\dfrac{9}{10}\)
\(\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-\frac{1}{30}=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}-\frac{1}{3\cdot4}-\frac{1}{4\cdot5}-\frac{1}{5\cdot6}\)
\(=\frac{1}{1}-\frac{1}{2}-\frac{1}{2}-\frac{1}{3}-\frac{1}{3}-\frac{1}{4}-\frac{1}{4}-\frac{1}{5}-\frac{1}{5}-\frac{1}{6}\)
\(=\frac{1}{1}-\frac{1}{6}=\frac{5}{6}\)
NHỚ K MK NHA. CHÚC BẠN HỌC TỐT
1/2 - 1/6 - 1/12 - 1/20 - 1/30
=1/1x2 - 1/2x3- 1/3x4 - 1/4x5 - 1/5x6
=1-1/2 + 1/2-1/3 + 1/3-1/4 + 1/4-1/5 +1/5-1/6
=1-1/6
=5/6
* Bạn tham khảo nhé *
1212 ++ 1616 ++ 112112 ++ 120120 ++ 130130 ++ 142142 ++ 156156
== 11×211×2 ++ 12×312×3 ++ 13×413×4 ++ 14×514×5 ++ 15×615×6 ++ 16×716×7 ++ 17×817×8
== 1111 −− 1212 ++ 1212 −− 1313 ++ 1313 −− 1414 ++ 1414 −− 1515 ++ 1515 −− 1616 ++ 1616 −− 1717 ++ 1717 −− 1818
== 1111 −− 1818
== 8888 −− 1818
== 78
Bài 1 :
a) Hai phân số có chung tử số thì ta so sánh mẫu nếu mẫu lớn hơn thì phân số đó bé hơn
Áp dụng vào đó ta có : 71 < 72 => 15/71 > 15/72
b) Ta có : 21/42 = 1/2 = 23/46
Áp dụng câu a ta có : 46 > 45 => 21/42 < 23/45
c) Ta có : 47/45 = 1 + 2/45 ; 48/46 = 1 + 2/46
Vì 2/45 > 2/46 => 47/45 > 48/46
d) Ta có : 1 - 13/25 = 12/25
1/3 = 12/36
Vì 12/25 > 12/36 => 13/25 > 3/7
Bài 2 :
D = 1/2 + 1/6 + 1/12 + 1/20 + ... + 1/110
D = 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + .... + 1/10.11
D = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + .... + 1/10 - 1/11
D = 1 - 1/11
D = 10/11
1/3+1/4+1/5+8/10+15/20+20/30
= 1/3+1/4+1/5+4/5+3/4+2/3
= (1/3+2/3)+(1/4+3/4)+(1/5+4/5)
= 1 + 1 + 1
=3
\(\dfrac{1}{10}+\dfrac{4}{20}+\dfrac{9}{30}+\dfrac{16}{40}+\dfrac{25}{50}+\dfrac{36}{60}+\dfrac{49}{70}+\dfrac{64}{80}+\dfrac{81}{90}\)
\(=\dfrac{1}{10}+\dfrac{1}{5}+\dfrac{3}{10}+\dfrac{2}{5}+\dfrac{1}{2}+\dfrac{3}{5}+\dfrac{7}{10}+\dfrac{4}{5}+\dfrac{9}{10}\)
\(=\left(\dfrac{1}{10}+\dfrac{3}{10}+\dfrac{7}{10}+\dfrac{9}{10}\right)+\left(\dfrac{1}{5}+\dfrac{2}{5}+\dfrac{3}{5}+\dfrac{4}{5}\right)+\dfrac{1}{2}\)
\(=2+2+\dfrac{1}{2}\)
\(=4+\dfrac{1}{2}\)
\(=\dfrac{8}{2}+\dfrac{1}{2}=\dfrac{9}{2}\)
10+20+30+...+90
= (10+90) + (20+80) + (30+70) + (40+60) + 50
= 100 x 4 + 50
= 400 + 50
= 450
a) 15+45 .x= −715 b) 13+(34−𝑥)=(−0,5)2
c) (8x3+1)⋅(4x2−9)=0 d) 3x+2+2.3x−1=261
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{90}+\frac{1}{110}\)
\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+....+\frac{1}{9\cdot10}+\frac{1}{10\cdot11}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)
\(=\frac{1}{1}-\frac{1}{11}\)
\(=\frac{10}{11}\)
\(\frac{1}{2}+\frac{1}{3}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{90}\)\(+\frac{1}{110}\)
\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+...\) \(+\frac{1}{9\cdot10}\)\(+\frac{1}{10\cdot11}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\)\(\frac{1}{5}\)\(+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{9}-\frac{1}{10}\)\(+\frac{1}{10}-\frac{1}{11}\)
\(=1-\frac{1}{11}\)
\(=\frac{10}{11}\)
\(3\cdot\frac{1}{4}+\frac{1}{12}\cdot\frac{1}{4}+\frac{35}{12}\cdot\frac{1}{4}\)
\(=(3+\frac{1}{12}+\frac{35}{12})\cdot\frac{1}{4}\)
\(=(3+\frac{36}{12})\cdot\frac{1}{4}\)
\(=(3+3)\cdot\frac{1}{4}\)
\(=6\cdot\frac{1}{4}=\frac{6}{4}\)
\(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{72}+\dfrac{1}{90}\)
\(=\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+...+\dfrac{1}{8\times9}+\dfrac{1}{9\times10}\)
\(=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=\dfrac{1}{3}-\dfrac{1}{10}\)
`=7/30`
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