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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
`A = 3/4 xx 8/9 xx ... xx 99/100`
`= (1xx3)/(2xx2) xx (2xx4)/(3xx3) xx ... xx (9xx11)/(10xx10)`
`= (1xx2xx3xx ... xx 9)/(2xx3xx...xx10) xx (3xx4xx5xx...xx 11)/(2xx3xx4xx...xx 10)`
`= 1/10 xx 11`
`= 11/10`.
Ta có: `11/10 > 1`
`11/19 < 1`.
`=> A > 11/19`.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)...\left(1-\frac{1}{81}\right)\left(1-\frac{1}{100}\right)\)
\(B=\frac{3}{4}\cdot\frac{8}{9}\cdot...\cdot\frac{80}{81}\cdot\frac{99}{100}\)
\(B=\frac{1.3}{2.2}\cdot\frac{2.4}{3.3}\cdot...\cdot\frac{8.10}{9.9}\cdot\frac{9.11}{10.10}\)
\(B=\frac{\left(1\cdot2\cdot...\cdot8\cdot9\right).\left(3\cdot4\cdot...\cdot10\cdot11\right)}{\left(2\cdot3\cdot..\cdot9\cdot10\right).\left(2\cdot3\cdot...\cdot9\cdot10\right)}\)
\(B=\frac{1\cdot2\cdot...\cdot8\cdot9}{2\cdot3\cdot...\cdot9\cdot10}\cdot\frac{3\cdot4\cdot...\cdot10\cdot11}{2\cdot3\cdot...\cdot9\cdot10}\)
\(B=\frac{1}{10}\cdot\frac{11}{2}=\frac{11}{20}\)
Vì 20 < 21 nên 11/20 > 11/21
Vậy .....
bạn vào link này nè:https://olm.vn/hoi-dap/question/980572.html
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: A = \(\frac{-2}{11}+\frac{6}{7}+\frac{1}{2}+\frac{-9}{11}+\frac{1}{7}\)
A = \(\left(\frac{-2}{11}+\frac{-9}{11}\right)+\left(\frac{6}{7}+\frac{1}{7}\right)+\frac{1}{2}\)
A = \(-1+1+\frac{1}{2}\)
A = \(\frac{1}{2}\)
B = \(\left(\frac{9}{16}+\frac{8}{27}\right)+\left(1+\frac{7}{16}+\frac{-19}{27}\right)\)
B = \(\frac{9}{16}+\frac{8}{27}+1+\frac{7}{16}-\frac{19}{27}\)
B = \(\left(\frac{9}{16}+\frac{7}{16}\right)+1+\left(\frac{8}{27}-\frac{19}{27}\right)\)
B = \(1+1-\frac{11}{27}\)
B = \(\frac{43}{27}\)
Mà 1/2 < 43/27 (Vì 1/2 < 1; 43/27 > 1)
=> A < B
Giải
\(A=\frac{-2}{11}+\frac{6}{7}+\frac{1}{2}+\frac{-9}{11}+\frac{1}{7}\)
\(\Leftrightarrow A=\left(\frac{-2}{11}+\frac{-9}{11}\right)+\left(\frac{6}{7}+\frac{1}{7}\right)+\frac{1}{2}\)
\(\Leftrightarrow A=\frac{-11}{11}+\frac{7}{7}+\frac{1}{2}\)
\(\Leftrightarrow A=-1+1+\frac{1}{2}\)
\(\Leftrightarrow A=\frac{1}{2}< 1\left(1\right)\)
\(B=\left(\frac{9}{16}+\frac{8}{27}\right)+\left(1+\frac{7}{16}+\frac{-19}{27}\right)\)
\(\Leftrightarrow B=\left(\frac{9}{16}+\frac{7}{16}\right)+\left(\frac{8}{27}+\frac{-19}{27}\right)+1\)
\(\Leftrightarrow B=\frac{16}{16}+\frac{-11}{27}+1\)
\(\Leftrightarrow B=1+\frac{-11}{27}+1\)
\(\Leftrightarrow B=2+\frac{-11}{27}\)
\(\Leftrightarrow B=\frac{43}{27}\)\(>1\left(2\right)\)
Từ (1) và (2) suy ra A < B
![](https://rs.olm.vn/images/avt/0.png?1311)
\(M=\frac{3}{4}\cdot\frac{8}{9}\cdot\cdot\cdot\frac{99}{100}=\frac{1\cdot3}{2\cdot2}\cdot\frac{2\cdot4}{3\cdot3}\cdot\cdot\cdot\frac{9\cdot11}{10\cdot10}=\frac{\left(1\cdot2\cdot\cdot\cdot\cdot9\right)\cdot\left(3\cdot4\cdot\cdot\cdot10\cdot11\right)}{\left(2\cdot3\cdot\cdot\cdot\cdot9\cdot10\right)\left(2\cdot3\cdot\cdot\cdot\cdot\cdot10\right)}=\frac{11}{2.10}=\frac{11}{20}\)\(
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:\(\frac{1}{11}>\frac{1}{20};\frac{1}{12}>\frac{1}{20};\frac{1}{13}>\frac{1}{20};....;\frac{1}{19}>\frac{1}{20}\)
\(\Rightarrow\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{20}>\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}\)(Có 10 phân số \(\frac{1}{20}\))
\(\Rightarrow\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}>\frac{10}{20}\)\(\Leftrightarrow S>\frac{10}{20}\)
Mà \(\frac{10}{20}=\frac{1}{2}\)nên
\(\Rightarrow S>\frac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{16}+\frac{1}{17}+\frac{1}{18}+\frac{1}{19}+\frac{1}{20}>\frac{1}{20}+...+\frac{1}{20}\)(10 số \(\frac{1}{20}\))
=\(\frac{1}{20}.10=\frac{1}{2}\)
vậy S>1/2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right)......\left(1-\frac{1}{81}\right)\left(1-\frac{1}{100}\right)\)
= \(-\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.......\frac{80}{81}.\frac{99}{100}\)
=\(-\frac{1.3.2.4.3.5..............8.10.9.11}{2^2.3^2.4^2.......10^2}=-\frac{\left(1.2.3.....9\right)\left(3.4.5....11\right)}{2.3.4....10.2.3.4.....10}=-\frac{11}{20}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 1619 và 825
Ta có :
1619 = ( 24 )19 = 276
825 = ( 23 )25 = 275
Vì 276 > 275 Nên 1619 > 825
b) 536 và 1124
Ta có :
536 = ( 53 )12 = 12512
1124 = ( 112 )12 = 12112
Vì 12512 > 12112 Nên 536 > 1124
1.
\(M=3^0+3^1+......+3^{50}.\)
\(\Rightarrow3M=3+3^2+.......+3^{51}\)
\(\Rightarrow3M-M=\left(3+3^2+.......+3^{51}\right)-\left(3^0+3+.....+3^{50}\right)\)
\(\Rightarrow2M=3^{51}-1\)
\(\Rightarrow M=\frac{3^{51}-1}{2}\)
2.
\(a,\)Ta có : \(16^{19}=\left(2^4\right)^{19}=2^{76}\)
\(8^{25}=\left(2^3\right)^5=2^{75}\)
Vì \(2^{76}>2^{75}\Rightarrow16^{19}>8^{25}\)
\(b,\)Ta có : \(5^{36}=\left(5^3\right)^{12}=125^{12}\)
\(11^{24}=\left(11^2\right)^{12}=121^{12}\)
Vì \(125^{12}>121^{12}\Rightarrow5^{36}>11^{24}\)
Đặt A= ( 1-\(\frac{1}{4}\)). ( 1-\(\frac{1}{9}\)).( 1-\(\frac{1}{16}\))......(1-\(\frac{1}{100}\))
Ta có:A= ( 1-\(\frac{1}{4}\)). ( 1-\(\frac{1}{9}\)).( 1-\(\frac{1}{16}\))......(1-\(\frac{1}{100}\))
A = \(\frac{3}{4}\).\(\frac{8}{9}\).\(\frac{15}{16}\).......\(\frac{99}{100}\)
A= \(\frac{1.3}{2.2}\). \(\frac{2.4}{3.3}\).\(\frac{3.5}{4.4}\).......\(\frac{9.11}{10.10}\)
A=\(\frac{1.2.3....9}{2.3.4....10}\).\(\frac{3.4.5....11}{2.3.4....10}\)
A= \(\frac{1}{10}\). \(\frac{11}{2}\)
A= \(\frac{11}{20}\)
Do 20> 19 => \(\frac{11}{20}\)< \(\frac{11}{19}\). Vậy A< \(\frac{11}{19}\)
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