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\(B=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)...\left(1-\frac{1}{81}\right)\left(1-\frac{1}{100}\right)\)
\(B=\frac{3}{4}\cdot\frac{8}{9}\cdot...\cdot\frac{80}{81}\cdot\frac{99}{100}\)
\(B=\frac{1.3}{2.2}\cdot\frac{2.4}{3.3}\cdot...\cdot\frac{8.10}{9.9}\cdot\frac{9.11}{10.10}\)
\(B=\frac{\left(1\cdot2\cdot...\cdot8\cdot9\right).\left(3\cdot4\cdot...\cdot10\cdot11\right)}{\left(2\cdot3\cdot..\cdot9\cdot10\right).\left(2\cdot3\cdot...\cdot9\cdot10\right)}\)
\(B=\frac{1\cdot2\cdot...\cdot8\cdot9}{2\cdot3\cdot...\cdot9\cdot10}\cdot\frac{3\cdot4\cdot...\cdot10\cdot11}{2\cdot3\cdot...\cdot9\cdot10}\)
\(B=\frac{1}{10}\cdot\frac{11}{2}=\frac{11}{20}\)
Vì 20 < 21 nên 11/20 > 11/21
Vậy .....
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a) (x - 3)(y - 3) = 9 = 1.9 = 3.3
Lập bảng:
| x - 3 | 1 | -1 | 3 | -3 | 9 | -9 |
| y - 3 | 9 | -9 | 3 | -3 | 1 | -1 |
| x | 4 | 2 | 6 | 0 | 12 | -3 |
| y | 12 | -6 | 6 | 0 | 4 | 2 |
Vậy ...
b) A = \(\frac{10^{19}+1}{10^{20}+1}\) => 10A = \(\frac{10^{20}+10}{10^{20}+1}=1+\frac{9}{10^{20}+1}\)
B = \(\frac{10^{20}+1}{10^{21}+1}\) => 10B = \(\frac{10^{21}+10}{10^{21}+1}=1+\frac{9}{10^{21}+1}\)
Do \(10^{20}+1< 10^{21}+1\) => \(\frac{9}{10^{20}+1}>\frac{9}{10^{21}+1}\) => 10A > 10B => A > B
Ta có: (1-1/2)(1-1/3)(1-1/4)......(1-1/20)
= (2/2-1/2)(3/3-1/3)(4/4-1/4)....(20/20-1/20)
= 1/2*2/3*3/4*...*19/20
=1/20
Vì 1/20>1/21
=> (1-1/2)(1-1/3)(1-1/4)......(1-1/20)>1/21
10A=\(\frac{10^{20}+10}{10^{20}+1}\)=\(\frac{10^{20}+1+9}{10^{20}+1}\)=\(1\)+\(\frac{9}{10^{20}+1}\)
10B=\(\frac{10^{21}+10}{10^{21}+1}\)=\(\frac{10^{21}+1+9}{10^{21}+1}\)=\(1\)+\(\frac{9}{10^{21}+1}\)
Vì \(\frac{9}{10^{20}+1}\)>\(\frac{9}{10^{21}+1}\)nên 10A>10B\(\Rightarrow\)A>B
1/ So sánh A với \(\frac{1}{4}\)
Có \(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+.........+\frac{1}{2014.2015.2016}\)
\(A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-.......+\frac{1}{2014.2015}-\frac{1}{2015.2016}\)
\(A=\frac{1}{1.2}-\frac{1}{2015.2016}=\frac{1}{2}-\frac{1}{2015.2016}\)
Vậy \(A>\frac{1}{4}\)
Ta thấy:
\(\frac{1}{2^2}<\frac{1}{1.2}\)
\(\frac{1}{3^2}<\frac{1}{2.3}\)
................
\(\frac{1}{19^2}<\frac{1}{18.19}\)
Cộng vế với vế ta có:
\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{19^2}<\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{18.19}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{18}-\frac{1}{19}\)\(=1-\frac{1}{19}=\frac{18}{19}>\frac{18}{40}=\frac{9}{20}\)
Kết luận: ....>.....
\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{20}\)
\(\Rightarrow\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{20}>\frac{1}{20}+\frac{1}{20}+..+\frac{1}{20}\left(19SH\right)\)
\(\Rightarrow\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+..+\frac{1}{20}>\frac{19}{20}\)
Vậy ................
Đặt \(A=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{20}\) ta có :
\(A>\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}\)
Do có \(20-2+1=19\) phân số \(\frac{1}{20}\) nên :
\(A>19.\frac{1}{20}=\frac{19}{20}\)
Vậy \(A>\frac{19}{20}\)
Chúc bạn học tốt ~
Ta có :
\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{19}\right).\left(1-\frac{1}{20}\right)\)
\(=\)\(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{18}{19}.\frac{19}{20}\)
\(=\)\(\frac{1.2.3.....18.19}{2.3.4.....19.20}\)
\(=\)\(\frac{1}{20}\)
Vì \(\frac{1}{20}>\frac{1}{21}\)nên \(A>\frac{1}{21}\)
Vậy \(A>\frac{1}{21}\)