Ta có: 1/4+1/5+...+1/10>1/10.7=7/10

1/11+1/12+...+1/19>1/20.9=9/20

Kết hợp lại ta có B= 1/4+1/5+1/6+...+1/19>7/10+9/20=23/20>1.Vậy B>1

ta co 1/4+1/5+......+1/10>1/10.7=7/10

1/11+1/12+.....1/19>1/20.9=9/20

kết hợp lại ta có mB=1/4+1/5+1/6+......1/19>7/10+9/20=23/20>1 vậy B>1

ta có:

1/4>1/10

1/5>1/10

1/6>1/10

1/7>1/10

1/8>1/10

1/9>1/10

=>1/4 +1/5+1/6+1/7+1/8+1/9>1/10*6

=>1/4+1/5+1/6+1/7+1/8+1/9+1/10>1/10*7

=>1/4+1/5+1/6+...+1/19>1/10*7+1/11+1/12+...+1/19

=>1/4+1/5+1/6+...+1/19>7/10+1/11+1/12+...+1/19

mà ta thấy:

1/11>1/20

1/12>1/20

1/13>1/20

1/14>1/20

1/15>1/20

...

1/19>1/20

=>1/11+1/12+1/13+...+1/19>1/20*9

=>1/4+1/5+1/6+...+1/19>1/10*7+1/20*9

=>B>7/10+9/20

=>B>23/20=1.15>1

=>B>1(dpcm)

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 \(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{19}>\frac{1}{19}+...+\frac{1}{19}=10\cdot\frac{1}{19}=\frac{10}{19}>\frac{1}{2}\)

\(\Rightarrow B>\frac{1}{4}+\frac{5}{9}+\frac{10}{19}>\frac{1}{4}+\frac{1}{2}+\frac{1}{2}=\frac{1}{4}+\frac{2}{4}+\frac{2}{4}\)

\(\Rightarrow B>\frac{5}{4}>1\Rightarrow B>1\)

Ta có:\(\frac{1}{5}>\frac{1}{6}>...>\frac{1}{19}\)

=>B=\(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{19}>\frac{1}{4}+\frac{1}{19}+\frac{1}{19}+...+\frac{1}{19}\)(15 lần 1/19)

=>B>1/4+15/19=79/76>1

=>B>1

Nhanh cc ! ngu đừng hỏi lắm => càng hỏi càng ngu vvvv

Ta có : A= 1/2^2 +1/3^2 +....+1/2012^2 +1/2013^2

=> A= 1/2.2 +1/3.3 +....+1/2012.2012  +1/2013.2013

Do :1/2.2< 1/1.2

      1/3.3 <1/2.3

       .................

       1/2012.2012 <1/2011.2012

       1/2013.2013< 1/2012.2013

=>1/2.2 +1/3.3 +...+1/2012.2012+1/2013.2013< 1/1.2 +1/2.3+...+1/2011.2012+1/2012.2013

=>A<1/1 -1/2 +1/2 -1/3+...+1/2011-1/2012+1/2012-1/2013

=>A<1/1-1/2013

=>A<2013/2013 -1/2013

=> A< 2012/2013

Vì 2012<2013=>2012/2013<1

mà A<2012/2013=>A<1

Vậy A<1

Ta có :

\(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{2012^2}< \frac{1}{2011.2012};\frac{1}{2013^2}< \frac{1}{2012.2013}\)

\(\Rightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2011.2012}+\frac{1}{2012.2013}\)

\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2011}-\frac{1}{2012}+\frac{1}{2012}-\frac{1}{2013}\)

\(\Rightarrow A< 1-\frac{1}{2013}\)

Mặt khác : \(1-\frac{1}{2013}< 1\)

\(\Rightarrow A< 1-\frac{1}{2013}< 1\)

\(\Rightarrow A< 1\)

\(\)

Ta có : B = 1/3 + 1/3^2 + 1/3^3 +...+ 1/3^99

=> 3B - B = ( 1 + 1/3 + 1/3^2 +...+ 1.3^99) - ( 1/3 + 1/3^2 + 1/3^3 +...+ 1/3^99 )

=> 2B = 1 - 1/3^99 < 1

=> 2B < 1

=> B < 1/2 ( ĐPCM )

\(B< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{7.8}\)

\(B< \frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{8-7}{7.8}\)

\(B< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{7}-\frac{1}{8}\)

\(B< 1-\frac{1}{8}< 1\left(dpcm\right)\)

\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{43}-\frac{1}{46}\)

\(s=1-\frac{1}{46}< 1\)

Vậy S<1

\(S=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{43\cdot46}\)

\(S=1\left[\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+...+\frac{1}{43\cdot46}\right]\)

\(S=1\left[1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{43}-\frac{1}{46}\right]\)

\(S=1\left[1-\frac{1}{46}\right]=1\cdot\frac{45}{46}=\frac{45}{46}< 1(đpcm)\)

 S = \(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)

S = \(\frac{1}{1}-\frac{1}{46}\)

S = \(\frac{46}{46}-\frac{1}{46}\)

S = \(\frac{45}{46}< 1\)

Hay S < 1

Vậy S < 1 (đpcm)