So sánh: A = (3 + 1)(32 + 1)(34 + 1)(38 + 1)(316 + 1) với B = 332 - 1
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a) \(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)=\dfrac{1}{2}\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)=\dfrac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)=\dfrac{1}{2}\left(3^{32}-1\right)< 3^{32}-1=B\)
b) \(A=2011.2013=\left(2012-1\right)\left(2012+1\right)=2012^2-1< 2012^2=B\)
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a) Ta có : 2005.2007 = (2006 - 1)(2006 + 1) = 20062 - 12 = 20062 - 1 ( cái khúc này sửa : 2005.2001 thành 2005.2007)
Mà B = 20062
=> 20062 - 1 < 20062
=> A < B
b) Ta có : B = (2 + 1)(22 + 1)(24 + 1)(28 + 1)(216 + 1)
B = (2 - 1)(2 + 1)(22 + 1)(24 + 1)(28 + 1)(216 + 1)
B = (22 - 1)(22 + 1)(24 + 1)(28 + 1)(216 + 1)
B = (24 - 1)(24 + 1)(28 + 1)(216 + 1)
B = (28 - 1)(28 + 1)(216 + 1) = (216 - 1)(216 + 1) = 232 - 1
Mà C = 232
=> B < C
c) Tương tự như câu b
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)=\dfrac{1}{2}\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)=\dfrac{1}{2}.\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)=\dfrac{1}{2}\left(3^{32}-1\right)=\dfrac{3^{32}}{2}-\dfrac{1}{2}\)
\(\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\dfrac{\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\)
\(=\dfrac{3^{32}-1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(4\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\dfrac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\dfrac{1}{2}\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\dfrac{1}{2}\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\dfrac{1}{2}\left(3^{16}-1\right)\cdot\left(3^{16}+1\right)\)
\(=\dfrac{1}{2}\left(3^{32}-1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\dfrac{1}{4}=\dfrac{10}{40}=\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}\)
Mà \(\dfrac{1}{31}>\dfrac{1}{40}\)
\(\dfrac{1}{32}>\dfrac{1}{40}\)
\(\dfrac{1}{33}>\dfrac{1}{40}\)
\(\dfrac{1}{34}>\dfrac{1}{40}\)
\(\dfrac{1}{35}>\dfrac{1}{40}\)
\(\dfrac{1}{36}>\dfrac{1}{40}\)
\(\dfrac{1}{37}>\dfrac{1}{40}\)
\(\dfrac{1}{38}>\dfrac{1}{40}\)
\(\dfrac{1}{39}>\dfrac{1}{40}\)
\(\Rightarrow\) \(\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+...+\dfrac{1}{39}+\dfrac{1}{40}>\dfrac{10}{40}=\dfrac{1}{4}\)
Vậy \(S>\dfrac{1}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=8.\left(3^2+1\right)\left(3^4+1\right)....\left(3^{16}+1\right)\\ =\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)....\left(3^{16}+1\right)\\ =\left(3^4-1\right)\left(3^4+1\right)....\left(3^{16}+1\right)\\ =\left(3^8-1\right)....\left(3^{16}+1\right)\\ =\left(3^{16}-1\right)\left(3^{16}+1\right)\\ =3^{32}-1\)
A = 8.(3² + 1)(3⁴ + 1)(3⁸ + 1)(3¹⁶ + 1)
= (3² - 1)(3² + 1)(3⁴ + 1)(3⁸ + 1)(3¹⁶ + 1)
= (3⁴ - 1)(3⁴ + 1)(3⁸ + 1)(3¹⁶ + 1)
= (3⁸ - 1)(3⁸ + 1)(3¹⁶ + 1)
= (3¹⁶ - 1)(3¹⁶ + 1)
= 3³² - 1
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
2)Ta có: \(2^{332}< 2^{333}=\left(2^3\right)^{111}=8^{111}\)
\(3^{223}>3^{222}=\left(3^2\right)^{111}=9^{111}\)
Vì \(8^{111}< 9^{111}\) mà \(2^{332}< 8^{111},3^{223}>9^{111}\) nên suy ra \(2^{332}< 3^{223}\)
Vậy \(2^{332}< 3^{223}\)
1) \(A=\dfrac{10^{2013}+1}{10^{2014}+1}\Rightarrow10A=\dfrac{10^{2014}+10}{10^{2014}+1}=\dfrac{10^{2014}+1}{10^{2014}+1}+\dfrac{9}{10^{2014}+1}=1+\dfrac{9}{10^{2014}+1}\)
\(B=\dfrac{10^{2014}+1}{10^{2015}+1}\Rightarrow10B=\dfrac{10^{2015}+10}{10^{2015}+1}=\dfrac{10^{2015}+1}{10^{2015}+1}+\dfrac{9}{10^{2015}+1}=1+\dfrac{9}{10^{2015}+1}\)Vì: \(10^{2014}+1< 10^{2015}+1\Rightarrow\dfrac{9}{10^{2014}+1}>\dfrac{9}{10^{2015}+1}\Rightarrow1+\dfrac{9}{10^{2014}+1}>1+\dfrac{9}{10^{2015}+1}\)
Nên suy ra \(10A>10B\Rightarrow A>B\)
Ta có:
\(3^{32}-1=\left(3^{16}-1\right)\left(3^{16}+1\right)\)
\(=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=2\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow A< B\)