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\(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{19}=\left(\frac{1}{4}+\frac{1}{5}+...+\frac{1}{11}\right)+\left(\frac{1}{12}+\frac{1}{13}+...+\frac{1}{19}\right)>\left(\frac{1}{11}+\frac{1}{11}+...+\frac{1}{11}\right)+\left(\frac{1}{19}+\frac{1}{19}+...+\frac{1}{19}\right)=\frac{8}{11}+\frac{8}{19}=\frac{240}{209}>\frac{209}{209}=1\Rightarrow B>1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{19}\)
\(B=\frac{1}{4}+\left(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{9}\right)+\left(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{19}\right)\)
\(\text{Mà:}\)\(\frac{1}{5}+\frac{1}{6}+...+1>\frac{1}{9}+\frac{1}{9}+...+\frac{1}{9}=\frac{5}{9}>\frac{1}{2}\)
\(\text{Mà:}\)\(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{19}>\frac{1}{19}+\frac{1}{19}+...+\frac{1}{19}=\frac{10}{19}>\frac{1}{2}\)
\(\text{Vậy:}\)\(B>1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
1/4>1/16 ; 1/5>1/16 ;1/6>1/16 ; ......; 1/19<1/16 (lấy phân số 1/16 vì từ 1/4 đến 1/19 có 16 số nên lấy 1/16 để được 1)
suy ra :
(1/4+1/5+1/6+...+1/15) >(1/16+1/16+1/16+...+1/16) =1 1/4+1/5+1/6+...1/15 >1 (1) (1/16+1/17+1/18+1/19) < (1/16+1/16+1/16+...+1/16) =1 1/16+1/17+1/18+1/19 <1 (2)
từ 1 và 2 suy ra b>1 là 11 lần (vì có 11 số) và b<1 là 4 lần (vì có 4 số) Vậy b>1
b có số số hạng là :
(19-4):1+1=16 ( số hạng)
16 chia hết cho 4 nên ta nhóm 4 số vào 1 nhóm
ta có B=(1/4+1/5+1/6+1/7)+(1/8+1/9+1/10+1/11)+(1/12+1/13+1/14+1/15)+(1/16+1/17+1/18+1/19)>(1/8+1/8+1/8+1/8)+(1/12+1/12+1/12+1/12)+(1/16+1/16+1/16+1/16)+(1/20+1/20+1/20+1/20)= 1/8.4+1/12.4+1/16.4+1/20.4=1/2+1/3+1/4+1/5=5/6+1/6=1
vậy b>1
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có B = 1/4 + 1/5 + 1/6+1/7+..+1/19
>1/19+1/19+..+1/19
=19/19=1
Vậy B>1
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{19}\)
\(=\frac{1}{4}+\left(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{9}\right)+\left(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{19}\right)\)
Ta có : \(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{9}>\frac{1}{9}+\frac{1}{9}+...+\frac{1}{9}=\frac{1}{9}.5=\frac{5}{9}>\frac{1}{2}\)
\(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{19}>\frac{1}{19}+\frac{1}{19}+...+\frac{1}{19}=\frac{1}{19}.5=\frac{5}{19}>\frac{1}{2}\)
\(\Rightarrowđpcm\)
![](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 :
\(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{9}>\frac{1}{9}+\frac{1}{9}+...+\frac{1}{9}=\frac{1}{9}.5=\frac{5}{9}>\frac{1}{2}\)
\(\frac{1}{10}+\frac{1}{11}+....+\frac{1}{19}>\frac{1}{19}+\frac{1}{19}+...+\frac{1}{19}=\frac{1}{19}.5>\frac{10}{19}>\frac{1}{2}\)
\(\Rightarrow B>\frac{1}{4}+\frac{1}{2}+\frac{1}{2}>1\)