cho B =\(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{19}\),hãy chứng minh B>1
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![](https://rs.olm.vn/images/avt/0.png?1311)
Bạn xem lời giải của mình nhé:
Giải:
Ta tách B làm 2 vế, mỗi vế có 8 số hạng:
+) \(A=\frac{1}{4}+\frac{1}{5}+...+\frac{1}{11}\)
+) \(C=\frac{1}{12}+\frac{1}{13}+...+\frac{1}{19}\)
Xét A:
1/4 > 1/12
1/5 > 1/12
...
1/11 > 1/12
=> \(\frac{1}{4}+\frac{1}{5}+...+\frac{1}{11}< \frac{1}{12}+\frac{1}{12}+...+\frac{1}{12}\) (8 số 1/12) => \(A< \frac{8}{12}\Rightarrow A< \frac{3}{4}\)(1)
Xét C:
1/12 > 1/20
1/13 > 1/20
...
1/19 > 1/20
=> \(\frac{1}{12}+\frac{1}{13}+...+\frac{1}{19}>\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}\) (8 số hạng) => \(C>\frac{8}{20}\Rightarrow C>\frac{2}{5}\)(2)
Từ (1) và (2) => A + C > \(\frac{3}{4}+\frac{2}{5}\Rightarrow B>1\frac{3}{20}>1\)
Vậy B>1 (đpcm)
Chúc bạn học tốt!
\(B=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{19}\)
\(B=\frac{1}{4}+\left(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{9}\right)+\left(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{19}\right)\)
Vì \(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{9}>\frac{1}{9}+\frac{1}{9}+...+\frac{1}{9}\) nên \(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{9}>\frac{5}{9}>\frac{1}{2}\)
Vì \(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{19}>\frac{1}{19}+\frac{1}{19}+...+\frac{1}{19}\) nên \(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{19}>\frac{10}{19}>\frac{1}{2}\)
\(=>\) \(B>\frac{1}{4}+\frac{1}{2}+\frac{1}{2}>1\)
Vậy \(B< 1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
\(B=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+..............+\frac{1}{19}\)
\(B=\frac{1}{4}+\left(\frac{1}{5}+\frac{1}{6}+.....+\frac{1}{9}\right)+\left(\frac{1}{10}+\frac{1}{11}+.........+\frac{1}{19}\right)\)
Ta thấy :
\(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{9}>\frac{1}{9}+\frac{1}{9}+...+\frac{1}{9}=\frac{1}{9}.5=\frac{5}{9}>\frac{1}{2}\)
\(\frac{1}{10}+\frac{1}{11}+....+\frac{1}{19}>\frac{1}{19}+\frac{1}{19}+...+\frac{1}{19}=\frac{1}{19}.5>\frac{10}{19}>\frac{1}{2}\)
\(\Rightarrow B>\frac{1}{4}+\frac{1}{2}+\frac{1}{2}>1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{19}=\frac{1}{4}+\frac{1}{17}+\frac{1}{18}+\frac{1}{19}+\left(\frac{1}{5}+...+\frac{1}{8}\right)+\left(\frac{1}{9}+...+\frac{1}{16}\right)\)
\(\frac{1}{5}+...+\frac{1}{8}>\frac{1}{8}.4=\frac{1}{2}\)
\(\frac{1}{9}+...+\frac{1}{16}\frac{1}{2}+\frac{1}{2}=1\)
\(SuyraB>1\)
Ta có: \(B=\left(\frac{1}{4}+\frac{1}{19}\right).8\)
\(B=2\frac{8}{19}\)
=> B>1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{19}>\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+...+\frac{1}{16}=\frac{16}{16}=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
3)
3/5 + 3/7-3/11 / 4/5 + 4/7- 4/11
= 3.( 1/5 + 1/7 - 1/11)/4.(1/5+1/7-1/11)
= 3/4
1,
ta có B = 196+197/197+198 = 196/(197+198) + 197/(197+198)
196/197 > 196/197+198
197/198 > 197/197+198
=> A>B
![](https://rs.olm.vn/images/avt/0.png?1311)
vì \(\frac{1}{4}< 1,\frac{1}{5}< 1,......,\frac{1}{19}< 1\) nên B < 1.
Ta có: \(B=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{19}\)
\(\Rightarrow B=\frac{1}{4}+\left(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{9}\right)+\left(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{19}\right)\)
Vì \(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{9}>\frac{1}{9}+\frac{1}{9}+...+\frac{1}{9}=5\cdot\frac{1}{9}=\frac{5}{9}>\frac{1}{2}\)
Vì \(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{19}>\frac{1}{19}+...+\frac{1}{19}=10\cdot\frac{1}{19}=\frac{10}{19}>\frac{1}{2}\)
\(\Rightarrow B>\frac{1}{4}+\frac{5}{9}+\frac{10}{19}>\frac{1}{4}+\frac{1}{2}+\frac{1}{2}=\frac{1}{4}+\frac{2}{4}+\frac{2}{4}\)
\(\Rightarrow B>\frac{5}{4}>1\Rightarrow B>1\)