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Nhận xét: \(\frac{1}{5}< \frac{1}{42};\frac{1}{9}< \frac{1}{42};\frac{1}{10}< \frac{1}{42};\frac{1}{40}< \frac{1}{42}\)
\(\Rightarrow S< \frac{1}{42}+\frac{1}{42}+\frac{1}{42}+\frac{1}{42}+\frac{1}{42}\)
\(\Rightarrow S< \frac{5}{42}< \frac{21}{42}=\frac{1}{2}\)
Vậy S < 1/2
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\(A=\frac{3^{10}+1}{3^9+1}=\frac{3^{10}+3-2}{3^9+1}=\frac{3\left(3^9+1\right)-2}{3^9+1}=3-\frac{2}{3^9+1}\)
\(B=\frac{3^9+1}{3^8+1}=\frac{3^9+3-2}{3^8+1}=\frac{3\left(3^8+1\right)-2}{3^8+1}=3-\frac{2}{3^8+1}\)
Có \(3^9+1>3^8+1\)
\(\Rightarrow\frac{2}{3^9+1}< \frac{2}{3^8+1}\)
\(\Rightarrow3-\frac{2}{3^9+1}>3-\frac{2}{3^8+1}\)
\(\Rightarrow A>B\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng bđt Cauchy cho 2 số dương \(\frac{a}{a+1}\)và\(\frac{a+1}{a}\)có
\(\frac{a}{a+1}+\frac{a+1}{a}\ge2\sqrt{\frac{a}{a+1}.\frac{a+1}{a}}=2\)
\(M=\frac{1}{1.2}+\frac{2}{1.2.3}+.....+\frac{9}{1.2.3.....10}\)
\(M=\frac{2-1}{1.2}+\frac{3-1}{1.2.3}+....+\frac{10-1}{1.2......10}\)
\(M=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{6}+....+\frac{10}{1.2.....10}-\frac{1}{1.2.....10}\)
\(M=1-\frac{1}{1.2.3......10}\)
\(M=1-\frac{1}{3628800}\)
Vì \(1=1\)mà \(\frac{1}{3628800}< 1\)nên \(1-\frac{1}{3628800}< 1\)
Vậy \(M< 1\)