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Ta có:5/20>5/25
5/21>5/25
5/22>5/25
5/23>5/25
5/24>5/25
=>S=5/20+5/21+5/22+5/23+5/24>5/25+5/25+5/25+5/25+5/25=1
=>5/20+5/21+5/22+5/23+5/24>1
DỄ
DO: 5/20 <1
5/21<1
5/22<1
5/23<1
5/24<1
=> 5/20+5/21+5/22+5/23+5/24<1
hay S<1 ( ĐPCM)
ĐÚNG NÈ ỦNG HỘ
CM: \(\dfrac{1\cdot3\cdot5\cdot7\cdot\cdot\cdot39}{21\cdot22\cdot23\cdot\cdot\cdot40}=\dfrac{1}{2^{20}}\)
Biến đổi vế trái:
\(\dfrac{1\cdot3\cdot5\cdot7\cdot\cdot\cdot39}{21\cdot22\cdot23\cdot\cdot\cdot40}=\dfrac{1\cdot3\cdot5\cdot7\cdot\cdot\cdot19}{22\cdot24\cdot26\cdot\cdot\cdot40}\)
\(=\dfrac{1\cdot3\cdot5\cdot7\cdot\cdot\cdot19}{2\cdot11\cdot2^3\cdot3\cdot2\cdot13\cdot2^2\cdot7\cdot2\cdot15\cdot2^5\cdot2\cdot17\cdot2^2\cdot9\cdot2\cdot19\cdot2^3\cdot5}\)
\(=\dfrac{1\cdot3\cdot5\cdot7\cdot\cdot\cdot19}{\left(3\cdot5\cdot7\cdot\cdot\cdot19\right)2^{20}}\)
\(=\dfrac{1}{2^{20}}\)
Ta có:\(\dfrac{1}{20}>\dfrac{1}{21}>\dfrac{1}{22}>\dfrac{1}{23}>\dfrac{1}{24}>\dfrac{1}{25}\)
=>S=\(\dfrac{5}{20}+\dfrac{5}{21}+\dfrac{5}{22}+\dfrac{5}{23}+\dfrac{5}{24}=5\left(\dfrac{1}{20}+\dfrac{1}{21}+\dfrac{1}{22}+\dfrac{1}{23}+\dfrac{1}{24}\right)>5\cdot\left(\dfrac{1}{25}+\dfrac{1}{25}+\dfrac{1}{25}+\dfrac{1}{25}+\dfrac{1}{25}\right)\)
=>S>\(5\cdot\dfrac{5}{25}\)
=>S>1(đpcm)
\(B=\left(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{150}\right)+\left(\dfrac{1}{151}+...\dfrac{1}{200}\right)>\dfrac{1}{150}+..\dfrac{1}{150}+\dfrac{1}{200}+..+200=\dfrac{50}{150}+\dfrac{50}{200}=\dfrac{1}{3}+\dfrac{1}{4}=\dfrac{4}{12}+\dfrac{3}{12}=\dfrac{7}{12}\)Vậy ... (ta có điều phải chứng minh )
Ta có :\(\dfrac{1}{20}>\dfrac{1}{200}\)
...
\(\dfrac{1}{199}>\dfrac{1}{200}\)
Do đó : \(\dfrac{1}{20}+\dfrac{1}{21}+...+\dfrac{1}{200}>\dfrac{1}{200}+\dfrac{1}{200}+..+\dfrac{1}{200}=\dfrac{181}{200}>\dfrac{180}{200}=\dfrac{9}{10}\)Vậy ...
a) \(5\dfrac{4}{23}.27\dfrac{3}{47}+4\dfrac{3}{47}.\left(-5\dfrac{4}{23}\right)\)
\(=5\dfrac{4}{23}.27\dfrac{3}{47}+\left(-4\dfrac{3}{47}\right).5\dfrac{4}{23}\)
\(=5\dfrac{4}{23}.\left[27\dfrac{3}{47}+\left(-4\dfrac{3}{47}\right)\right]\)
\(=5\dfrac{4}{23}.\left(27\dfrac{3}{47}-4\dfrac{3}{27}\right)\)
\(=5\dfrac{4}{23}.23\)
\(=\dfrac{119}{23}.23\)
\(=\dfrac{119}{23}\)
b) \(4.\left(\dfrac{-1}{2}\right)^3+\dfrac{3}{2}\)
\(=4.\dfrac{-1}{6}+\dfrac{3}{2}\)
\(=\dfrac{-4}{6}+\dfrac{3}{2}\)
\(=\dfrac{-2}{3}+\dfrac{3}{2}\)
\(=\dfrac{-4}{6}+\dfrac{9}{6}\)
\(=\dfrac{5}{6}\)
c) \(\left(\dfrac{1999}{2011}-\dfrac{2011}{1999}\right)-\left(\dfrac{-12}{1999}-\dfrac{12}{2011}\right)\)
\(=\dfrac{1999}{2011}-\dfrac{2011}{1999}-\dfrac{-12}{1999}+\dfrac{12}{2011}\)
\(=\left(\dfrac{1999}{2011}+\dfrac{12}{2011}\right)-\left(\dfrac{2011}{1999}+\dfrac{-12}{1999}\right)\)
\(=\dfrac{2011}{2011}-\dfrac{1999}{1999}\)
\(=1-1\)
\(=0\)
d) \(\left(\dfrac{-5}{11}+\dfrac{7}{22}-\dfrac{-4}{33}-\dfrac{5}{44}\right):\left(\dfrac{381}{22}-39\dfrac{7}{22}\right)\)
(đợi đã, mình chưa tìm được hướng làm...)
\(\left(\dfrac{1}{5}-\dfrac{1}{6}-\dfrac{1}{30}\right).\left(\dfrac{21}{22}+\dfrac{22}{23}+...+\dfrac{102}{103}\right)\)
\(=0.\left(\dfrac{21}{22}+\dfrac{22}{23}+...+\dfrac{102}{103}\right)\)
\(=0\)
vì \(\dfrac{1}{5}-\dfrac{1}{6}-\dfrac{1}{30}\)=0 nên
(15−16−13015−16−130 ).(2122+2223+......+102103)=0