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\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{14.15.16}\)
\(=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{14.15.16}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{14.15}-\frac{1}{15.16}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{15.16}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{240}\right)\)
\(=\frac{1}{2}.\frac{119}{240}\)
\(=\frac{119}{480}\)
Bài làm:
Ta có:\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{14.15.16}\)
\(=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{14.15.16}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{14.15}-\frac{1}{15.16}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{15.16}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{240}\right)\)
\(=\frac{1}{2}.\frac{119}{240}=\frac{119}{480}\)
\(\Rightarrow\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\right).y=\frac{49}{100}\)
\(\Leftrightarrow\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{100-98}{98.99.100}\right).y=\frac{49}{100}\)
\(\Leftrightarrow\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\right).y=\frac{49}{100}\)
\(\Leftrightarrow\left(\frac{1}{1.2}-\frac{1}{99.100}\right).y=\frac{49}{100}\Leftrightarrow\left(\frac{99.50-1}{99.100}\right).y=\frac{49}{100}\)
\(\Leftrightarrow\left(\frac{99.50-1}{99}\right).y=49\Leftrightarrow\left(99.50-1\right).y=99.49\Rightarrow y=\frac{99.49}{99.50-1}\)
Ta có : \(N=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{1000.1001}\)
\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{1001-1000}{1000.1001}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1000}-\frac{1}{1001}\)
\(=1-\frac{1}{1001}=\frac{1000}{1001}\)
Ta thấy : \(1001< 2020\Rightarrow\frac{1}{1001}>\frac{1}{2020}\)
\(\Rightarrow-\frac{1}{1001}< -\frac{1}{2020}\)
\(\Rightarrow1-\frac{1}{1001}< 1-\frac{1}{2020}\Rightarrow\frac{1000}{1001}< \frac{2019}{2020}\)
Hay : \(N< M\)
\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right).x=\frac{23}{45}\)
\(\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\right).x=\frac{23}{45}\)
\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right).x=\frac{23}{45}\)
\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{9.10}\right).x=\frac{23}{45}\)
\(\frac{1}{2}.\frac{22}{45}.x=\frac{23}{45}\)
\(\frac{11}{45}.x=\frac{23}{45}\)
\(x=\frac{23}{45}:\frac{11}{45}\)
\(x=\frac{23}{11}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\)
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{5}+\frac{1}{6}\)
=\(1-\frac{1}{6}\)
=\(\frac{5}{6}\)
Bài này mình chắc 100%, 1 đúng nha vì ghi cực khổ lắm:
1) Ta có: \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}...+\frac{50-49}{49.50}\)
\(=\frac{2}{1.2}-\frac{1}{1.2}+\frac{3}{2.3}-\frac{2}{2.3}+...+\frac{50}{49.50}-\frac{49}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}=1-\frac{1}{50}<1\)
2) Tương tự: \(S=\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\frac{1}{2}-\frac{1}{50}=\frac{24}{50}\)
\(\left(1\cdot2\right)^{-1}+\left(2\cdot3\right)^{-1}+\cdot\cdot\cdot+\left(9\cdot10\right)^{-1}\)
\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdot\cdot\cdot+\frac{1}{9\cdot10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdot\cdot\cdot+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}\)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdot\cdot\cdot+\frac{1}{9\cdot10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdot\cdot\cdot+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}\)
Đề \(A=\frac{1.98+2.97+3.96+...+98.1}{1.2+2.3+3.4+...+98.99}\) chứ bn!
Bài làm
ta có: 1.98 +2.97 + 3.96 + 98.1
= 1 + (1+2) + ( 1+2+3) +...+ ( 1+2+3+...+ 98)
\(=\frac{1.2}{2}+\frac{2.3}{2}+\frac{3.4}{2}+...+\frac{98.99}{2}\)
\(=\frac{1.2+2.3+3.4+...+98.99}{2}\)
\(\Rightarrow A=\frac{1.98+2.99+3.96+...+98.1}{1.2+2.3+3.4+...+98.99}\)
\(A=\frac{\left(1.2+2.3+3.4+...+98.99\right).\frac{1}{2}}{1.2+2.3+3.4+...+98.99}\)
\(A=\frac{1}{2}\left(đpcm\right)\)
Câu 1:
\(C=\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)+\left(1+\frac{1}{3.5}\right)+...\left(1+\frac{1}{2014.2016}\right)\)
\(\Rightarrow C=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}....\frac{2015.2015}{2014.2016}\)
\(\Rightarrow C=\frac{4.9.16...2015.2015}{3.8.15...2014.2016}\)
\(\Rightarrow C=\frac{2.2.3.3.4.4...2015.2015}{1.3.2.4...2014.2016}\)
\(\Rightarrow C=\frac{2.3.4...2015.2.3.4...2015}{1.2.3...2014.3.4.5...2016}\)
\(\Rightarrow C=\frac{2015}{1008}.\)
Vậy \(C=\frac{2015}{1008}.\)
Câu 2:
Do p là số nguyên tố lớn hơn 3 nên p có dạng \(3k+1\)hoặc\(3k+2\)
+ Nếu \(p=3k+1\Rightarrow p^2-1=\left(3k+1\right)^2-1\)
\(=9k^2+3k+3k+1-1\)
\(=9k^2+6k⋮3.\)( 1 )
+ Nếu \(p=3k+2\Rightarrow p^2-1=\left(3k+2\right)^2-1\)
\(=9k^2+6k+6k+4-1\)
\(=9k^2+12k+3⋮3\)( 2 )
Từ ( 1 ) và ( 2 ) \(\Rightarrow p^2-1⋮3\left(đpcm\right).\)
Câu 3:
\(2^{100}=\left(2^{10}\right)^{10}=1024^{10}>1000^{10}=10^{30}.\)( 1 )
\(2^{100}=2^{31}.2^6.2^{63}=2^{31}.64.512^7< 2^{31}.125.625^7=2^{31}.5^{31}=\)\(10^{31}.\)( 2 )
Từ ( 1 ) và ( 2 ) \(\Rightarrow10^{30}< 2^{100}< 10^{31}.\)
\(\Rightarrow\)2100 khi viết trong hệ thập phân có 31 chữ số.
Đáp số: 31 chữ số.
Câu 1 :
C = (1 + 1/1.3)(1 + 1/2.4)(1 + 1/3.5) .... (1 + 1/2014.2016)
C = (1.3/1.3 + 1/1.3) (2.4/2.4 + 1/2.4) ... (2014.2016/2014.2016 + 1/2014.2016)
C = 2.2/1.3 * 3.3/2.4 * ... * 2015.2015/2014.2016
C = 2.3....2015/1.2....2014 * 2.3....2015/3.4....2016
C = 2015 * 1/1008
C = 2015/1008
2y= 2/ 1.2.3 + 2/2.3.4 + 2/3.4.5 +.... +2/998.999.1000
2y=1/1.2 - 1/2.3 +1/2.3 - 1/3.4 + 1/3.4 -1/4.5 +....+ 1/998.999 - 1/ 999.1000
2y=1/2 - 1/ 999.1000
2y = 499500-1 / 999.1000
2y=499499 / 999.1000
y=499499 / 1998000
Ủng hộ mk nha
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