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\(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+\frac{4}{5}+\frac{5}{6}\) <4
=\(\frac{71}{20}\) <4
\(\Rightarrow\)3.55 < 4( đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
1.
Ta có:
1/2 < 2/3
3/4 < 4/5
.............
99/100 < 100/101
=> 1/2*3/4*5/6*...*99/100 < 2/3*4/5*6/7*...*100/101
=> A < B
2.
\(A\cdot B=\left[\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{99}{100}\right]\cdot\left[\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot...\cdot\frac{100}{101}\right]\)
\(A\cdot B=\frac{\left[1\cdot3\cdot5\cdot7\cdot...\cdot99\right]\left[2\cdot4\cdot6\cdot8\cdot...\cdot100\right]}{\left[2\cdot4\cdot6\cdot8\cdot...\cdot100\right]\left[3\cdot5\cdot7\cdot9\cdot...\cdot101\right]}=\frac{1\cdot3\cdot5\cdot...\cdot99}{3\cdot5\cdot7\cdot...\cdot101}=\frac{1}{101}\)
3.
Vì A < B => A.A < A.B => A2 < 1/101 < 1/100
Mà A2 < 1/100 <=> A2 < \(\frac{1}{10}^2\)=> A < 1/10
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 3:
\(A=5+5^2+..+5^{12}\)
\(5A=5\cdot\left(5+5^2+..5^{12}\right)\)
\(5A=5^2+5^3+...+5^{13}\)
\(5A-A=\left(5^2+5^3+...+5^{13}\right)-\left(5+5^2+...+5^{12}\right)\)
\(4A=5^2+5^3+...+5^{13}-5-5^2-...-5^{12}\)
\(4A=5^{13}-5\)
\(A=\dfrac{5^{13}-5}{4}\)
cho:
m = 1/2*3/4*5/6*....*99/100
n = 2/3*4/5*6/7*...*100/101
a, Chứng tỏ m<n
b,Tìm m*n
c, chứng tỏ m<1/10
![](https://rs.olm.vn/images/avt/0.png?1311)