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![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\dfrac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\)
\(=\dfrac{2^{10}\cdot3^8-2\cdot2^9\cdot3^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot2^2\cdot5}\)
\(=\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+2^{10}\cdot3^8\cdot5}\)
\(=\dfrac{2^{10}\cdot3^8\left(1-3\right)}{2^{10}\cdot3^8\left(1+5\right)}=\dfrac{-2}{6}=-\dfrac{1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A=(1/2^2-1) (1/3^2-1) (1/4^2-1) .... (1/100^2-1)
A=(1/2^2-2^2/2^2) (1/3^2-3^2/3^2) ...... (1/100^2-100^2/100^2)
A=1-2^2/2^2 . 1-3^2/3^2 .... 1-100^2/100^2
A=-(2^2-1/2^2 . 3^2-1/3^2 ..... 100^2-1/100^2)
A=-(1.3/2^2 x 2.4/3^2 ..... 99.101/100^2)
A=-(1.3.2.4.....99.101/2.2.3.3.....100.100)
A=-[(1.2.3....99).(3.4.5.....101) / (2.3.4...100) . (2.3.4...100) ]
A=-101/200 < -1/2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(M=1-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{10}}\right)\)
Đặt \(N=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{10}}\)
\(2N=1+\dfrac{1}{2}+...+\dfrac{1}{2^9}\)
\(\Rightarrow2N-N=1-\dfrac{1}{2^{10}}\)
\(\Rightarrow N=1-\dfrac{1}{2^{10}}\)
\(\Rightarrow M=1-\left(1-\dfrac{1}{2^{10}}\right)=\dfrac{1}{2^{10}}>\dfrac{1}{2^{11}}\)
Vậy \(M>\dfrac{1}{2^{11}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{n\left(n-1\right)}\\ A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}=1-\dfrac{1}{n}< 1\left(\dfrac{1}{n}>0\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bạn thiếu đề rồi phải là trừ hay cộng j j chứ.
Xét:
`A+B=2+1/2+1/3+1/4+......+1/4026+1/3+1/5+1/7+......+1/4025`
`1/2+1/3+1/4+......+1/4026+1/3+1/5+1/7+......+1/4025>0`
`=>A+B>2`
Mà `1 2013/2014<2`
`=>A+B>1 2013/2014`
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có:
2A=2.(12+122+123+...+122020+122021)2�=2.12+122+123+...+122 020+122 021
2A=1+12+122+123+...+122019+1220202�=1+12+122+123+...+122 019+122 020
Suy ra: 2A−A=(1+12+122+123+...+122019+122020)2�−�=1+12+122+123+...+122 019+122 020
−(12+122+123+...+122020+122021)−12+122+123+...+122 020+122 021
Do đó A=1−122021<1�=1−122021<1.
Lại có B=13+14+15+1360=20+15+12+1360=6060=1�=13+14+15+1360=20+15+12+1360=6060=1.
Vậy A < B.
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(A=\left(\dfrac{1}{2}-1\right)\cdot\left(\dfrac{1}{3}-1\right)\cdot\left(\dfrac{1}{4}-1\right)...\left(\dfrac{1}{10}-1\right)\)
\(A=-\dfrac{1}{2}\cdot-\dfrac{2}{3}-\dfrac{3}{4}\cdot...\cdot-\dfrac{9}{10}\)
\(A=\dfrac{-1\cdot-2\cdot-3\cdot...\cdot-9}{2\cdot3\cdot4\cdot...\cdot10}\)
\(A=-\dfrac{1}{10}\)
Mà: \(10>9\)
\(\Rightarrow\dfrac{1}{10}< \dfrac{1}{9}\)
\(\Rightarrow-\dfrac{1}{10}>-\dfrac{1}{9}\)
\(\Rightarrow A>-\dfrac{1}{9}\)