so sánh (1/3)^500va (1/5)^200
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Bài 1:
a: Sửa đề: 1/3^200
1/2^300=(1/8)^100
1/3^200=(1/9)^100
mà 1/8>1/9
nên 1/2^300>1/3^200
b: 1/5^199>1/5^200=1/25^100
1/3^300=1/27^100
mà 25^100<27^100
nên 1/5^199>1/3^300
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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\(\dfrac{1}{2^{500}}=\dfrac{1}{\left(2^5\right)^{100}}=\dfrac{1}{32^{100}}\\ \dfrac{1}{5^{200}}=\dfrac{1}{\left(5^2\right)^{100}}=\dfrac{1}{25^{100}}\)
mà `32^(100)>25^(100)`
nên \(\dfrac{1}{2^{500}}>\dfrac{1}{5^{200}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta đã biết \(\dfrac{1}{a\cdot a}< \dfrac{1}{\left(a+1\right)\left(a-1\right)}\) ( a ϵ Z )
⇒ \(Q=\dfrac{1}{2\cdot2}+\dfrac{1}{3\cdot3}+\dfrac{1}{4\cdot4}+...+\dfrac{1}{200\cdot200}\) < \(\dfrac{1}{1\cdot3}+\dfrac{1}{2\cdot4}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{199\cdot201}\)
Ta có \(\dfrac{1}{1\cdot3}+\dfrac{1}{2\cdot4}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{199\cdot201}\)
= \(\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{2\cdot4}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{199\cdot201}\right)\)
= \(\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{199}-\dfrac{1}{201}\right)\)
= \(\dfrac{1}{2}\left(1-\dfrac{1}{201}\right)=\dfrac{1}{2}\cdot\dfrac{200}{201}=\dfrac{100}{201}< \dfrac{100}{200}=\dfrac{1}{2}< \dfrac{3}{4}\)
Vậy Q < \(\dfrac{3}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
TC:(1/2)^300=(1/8)^100
(1/3)^200=(1/9)^100
Vì (1/8)^100>(1/9)^100 =>(1/2)^300 >(1/3)^200