Đặt 

Dễ thấy: 

Ta có:

Từ  ta có: 

Ta có : \(\frac{1}{32}+\frac{1}{42}+\frac{1}{52}+...+\frac{1}{102}< \frac{1}{32}+\frac{1}{32}+\frac{1}{32}+...+\frac{1}{32}\)   (8 số hạng)

\(\Rightarrow\frac{1}{32}+\frac{1}{42}+\frac{1}{52}+...+\frac{1}{102}< \frac{1}{32}.8=\frac{1}{4}< \frac{1}{2}\)

\(\Rightarrow\frac{1}{32}+\frac{1}{42}+\frac{1}{52}+...+\frac{1}{102}< \frac{1}{2}\left(đpcm\right)\)

\(A=\frac{1}{32}+\frac{1}{42}+...+\frac{1}{102}< \frac{1}{32}+\frac{1}{32}+...+\frac{1}{32}=\frac{8}{32}< \frac{16}{32}=\frac{1}{2}\)

Vậy \(A< \frac{1}{2}\)

a)\(\dfrac{1}{2^2}<\dfrac{1}{1.2}\)

\(\dfrac{1}{3^3}<\dfrac{1}{2.3}\)

\(...\)

\(\dfrac{1}{8^2}<\dfrac{1}{7.8}\)

Vậy ta có biểu thức:

\(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{8^2}<\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{7.8}\)

\(B= 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{7}-\dfrac{1}{8}\)

\(B<1-\dfrac{1}{8}=\dfrac{7}{8}<1\)

Vậy B < 1 (đpcm)

Giải:

a) Ta có:

1/2²=1/2.2 < 1/1.2

1/3²=1/3.3 < 1/2.3

1/4²=1/4.4 < 1/3.4

1/5²=1/5.5 < 1/4.5

1/6²=1/6.6 < 1/5.6

1/7²=1/7.7 < 1/6.7

1/8²=1/8.8 <1/7.8

⇒B<1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8

   B<1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8

   B<1/1-1/8

   B<7/8

mà 7/8<1

⇒B<7/8<1

⇒B<1

b)S=3/1.4+3/4.7+3/7.10+...+3/40.43+3/43.46

   S=1/1-1/4+1/4-1/7+1/7-1/10+...+1/40-1/43+1/43-1/46

   S=1/1-1/46

   S=45/46

Vì 45/46<1 nên S<1

Vậy S<1

Chúc bạn học tốt!

https://olm.vn/cau-hoi/a-cho-a12211216211002-ctr-a12-b-cho-p122132142120232-ctr-p-khong-la-so-tu-nhien-c-cho-c132152172120211.8293222842881

Cô làm rồi em nhá

Câu a, xem lại đề bài

Câu b: 

    P =  \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{4^2}\) + ...+ \(\dfrac{1}{2023^2}\)

   Vì  \(\dfrac{1}{2^2}\) < \(\dfrac{1}{1.2}\)                =  \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\)

         \(\dfrac{1}{3^2}\) < \(\dfrac{1}{2.3}\)                = \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\)

         \(\dfrac{1}{4^2}\)  < \(\dfrac{1}{3.4}\)               = \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) 

     ........................

        \(\dfrac{1}{2023^2}\) < \(\dfrac{1}{2022.2023}\) = \(\dfrac{1}{2022}\) - \(\dfrac{1}{2023}\)

Cộng vế với vế ta có:  

0< P < 1 - \(\dfrac{1}{2023}\) < 1

Vậy 0 < P < 1 nên P không phải là số tự nhiên vì không tồn tại số tự nhiên giữa hai số tự nhiên liên tiếp

Câu c:  

C = \(\dfrac{1}{3^2}\) + \(\dfrac{1}{5^2}\) + \(\dfrac{1}{7^2}\) + ....+ \(\dfrac{1}{2021^2}\) + \(\dfrac{1}{2023^2}\) = C 

B =  \(\dfrac{1}{2^2}\) + \(\dfrac{1}{4^2}\) + \(\dfrac{1}{6^2}\)+.......+ \(\dfrac{1}{2020^2}\) + \(\dfrac{1}{2023^2}\) > 0 

Cộng vế với vế ta có: 

C+B =  \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{4^2}\) + \(\dfrac{1}{5^2}\)+ \(\dfrac{1}{6^2}\)+...+ \(\dfrac{1}{2023^2}\) > C + 0 = C > 0

             Mặt khác ta có: 

1 > \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\)+...+ \(\dfrac{1}{2023^2}\) (cm ở ý b)

Vậy 1 > C > 0 hay C không phải là số tự nhiên (đpcm)

A = \(\dfrac{1}{5^2}\) + \(\dfrac{1}{6^2}\) + \(\dfrac{1}{7^2}\) +.................+ \(\dfrac{1}{2004^2}\)

A = \(\dfrac{1}{5.5}\) + \(\dfrac{1}{6.6}\) + \(\dfrac{1}{7.7}\)+..............+ \(\dfrac{1}{2004.2004}\)

Vì \(\dfrac{1}{5}>\dfrac{1}{6}>\dfrac{1}{7}>...........>\dfrac{1}{2004}\)

nên ta có : \(\dfrac{1}{5.5}>\dfrac{1}{5.6}>\dfrac{1}{6.6}>\dfrac{1}{6.7}>\dfrac{1}{7.7}>.....>\dfrac{1}{2004.2004}>\dfrac{1}{2004.2005}\)

\(\dfrac{1}{5.5}+\dfrac{1}{6.6}+\dfrac{1}{7.7}+...+\dfrac{1}{2004.2004}>\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+..+\dfrac{1}{2004.2005}\)

A > \(\dfrac{1}{5}\) \(-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+....+\dfrac{1}{2004}-\dfrac{1}{2005}\)

A > \(\dfrac{1}{5}\) - \(\dfrac{1}{2005}\) = \(\dfrac{1}{5}\) - \(\dfrac{12}{24060}\)

\(\dfrac{1}{65}\) = \(\dfrac{1}{5}\) - \(\dfrac{12}{65}\) 

Vì \(\dfrac{12}{65}\) > \(\dfrac{12}{24060}\) nên A>  \(\dfrac{1}{65}\) ( phân số nào có phần bù nhỏ hơn thì phân số đó lớn hơn)

Tương tự ta có :

A = \(\dfrac{1}{5.5}\) + \(\dfrac{1}{6.6}\)+ \(\dfrac{1}{7.7}\)+......+\(\dfrac{1}{2004.2004}\) >\(\dfrac{1}{4.5}\)+\(\dfrac{1}{5.6}\)+.....\(\dfrac{1}{2003.2004}\)

A < \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{6}\) +......+ \(\dfrac{1}{2003}\) - \(\dfrac{1}{2004}\)

A < \(\dfrac{1}{4}-\dfrac{1}{2004}\) < \(\dfrac{1}{4}\)

\(\dfrac{1}{65}< \)A < \(\dfrac{1}{4}\) (đpcm)

8: =>2x=210

hay x=105

cho a =1/2.3/4.5/6.....99/100.Chứng minh rằng:1/15<a<1/10.

ta co a < 2/3.4/5.....100/101 
nhan hai ve cho a ta co 
a^2 <2/3.4/5...100/101.1/2.3/4.5/6...99/100 
a^2<1/101 <1/100 
a< can 1/100 a <1/10.

Cm tương tự ta dc a>1/15.

Bn cx có thể kham khảo bài làm khác là:https://diendan.hocmai.vn/threads/toan-6-cmr-a-1-10-va-a-1-15.223994/

sao lại nhỏ hơn

Ta thấy \(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)

 \(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)

......

\(\dfrac{1}{10^2}< \dfrac{1}{9.10}\)

hay \(D=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+....+\dfrac{1}{10^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\)

\(D< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{9}-\dfrac{1}{10}\)

\(D< 1-\dfrac{1}{10}=\dfrac{9}{10}< 1\) ( đpcm )

Ta có \(\dfrac{1}{2.2}\) < \(\dfrac{1}{1.2}\)

         \(\dfrac{1}{3.3}\)<\(\dfrac{1}{2.3}\)

         \(\dfrac{1}{4.4}\)<\(\dfrac{1}{3.4}\)

  .........................

         \(\dfrac{1}{10.10}\)<\(\dfrac{1}{9.10}\)

=>\(\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{10.10}\)\(< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\)

=> D <  1 - \(\dfrac{1}{10}\)

=>D < \(\dfrac{9}{10}\)

=> D < \(\dfrac{10}{10}\)

 Vậy D < 1

Sửa đề: CM A>4/9

A=1/3^2+1/4^2+...+1/60^2

=>A>1/3*4+1/4*5+...+1/60*61

=>A>1/3-1/61=58/183>4/9