có: 1/3^2<1/2.3; 1/4^2<1/3.4:...: 1/100^2<1/99.100

Mà: 1/1.2+1/2.3+...+1/99.100=1-1/2+1/2-1/3+...+1/99-1/100

=1-1/100

=99/100

=> 1/3^2+1/4^2+...+1/100^2<99/100<1

=> đpcm

UNDERSTAND ???

đặt A= biểu thức trên

tao có 

A<1/2.3+1/3.4+...+1/99.100

A<1/2-1/3+1/3-1/4+...+1/99-1/100

A<1/2-1/100<1/2

SUY RA A<1/2(DPCM)

Ta có

\(P< \frac{1}{4.5}+\frac{1}{5.6}+......+\frac{1}{99.100}\)

\(\Rightarrow P< \frac{1}{4}-\frac{1}{5}+.....+\frac{1}{99}-\frac{1}{100}\)

\(\Rightarrow P< \frac{1}{4}-\frac{1}{100}< \frac{1}{4}\)

\(\Rightarrow P< \frac{1}{4}\left(1\right)\)

\(p>\frac{1}{5^2}+\frac{1}{6.7}+....+\frac{1}{100.101}\)

\(P>\frac{1}{5^2}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{101}\)

\(P>\frac{1}{6}+\frac{1}{25}-\frac{1}{101}\)

Ta thấy

\(\frac{1}{25}>\frac{1}{101}\Rightarrow\frac{1}{25}-\frac{1}{101}>0\)

Đặt \(M=\frac{1}{25}-\frac{1}{101}\)

\(\Rightarrow P>\frac{1}{6}+M>\frac{1}{6}\)

\(\Rightarrow P>\frac{1}{6}\left(2\right)\)

Tự (1) và (2)

\(\Rightarrow\frac{1}{6}< p< \frac{1}{4}\)

Hình như sai đề thì phải chứ mk làm ko đc !!!

  A < 1/(1.2) + 1/(2.3) + 1/(3.4) + ...+ 1/(99.100) 
<=> A< 1- 1/2 + 1/2 - 1/3 + 1/4 - 1/5 + .. + 1/99 - 1/100 
<=> A < 1 - 1/100 < 1 (đpcm) 

So với  thì đây

Ta có:

\(\frac{1}{5^2}<\frac{1}{4.5}\)

\(\frac{1}{6^2}<\frac{1}{5.6}\)

\(...\)

\(\frac{1}{100^2}<\frac{1}{99.100}\)

\(\Rightarrow\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}<\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)

\(\Rightarrow\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}<\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)

\(\Rightarrow\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}<\frac{1}{4}-\frac{1}{100}<\frac{1}{4}<\frac{1}{2}\)

Vậy \(\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}<\frac{1}{2}\)

\(s=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}\)

\(S=\frac{1}{5.5}+\frac{1}{6.6}+\frac{1}{7.7}+...+\frac{1}{100.100}<\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{100.101}\)

\(S<\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{100.101}=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{100}-\frac{1}{101}\)

\(\Rightarrow S<\frac{1}{5}-\frac{1}{101}\)

Vì \(\frac{1}{5}<\frac{1}{2}\)nên \(\frac{1}{5}-\frac{1}{101}<\frac{1}{2}\)

hay \(S=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}<\frac{1}{5}-\frac{1}{101}<\frac{1}{2}\)

Vậy \(S=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}<\frac{1}{2}\)  (đpcm)

\(∘backwin\)

\(a ) ( x + 1 ) + ( x + 2 ) + ( x + 3 ) + ... + ( x + 100 ) = 5750\)

\( ( x + x + x + ... + x ) + ( 1 + 2 + 3 + ... + 100 ) = 5750 \)

\( 100 x + ( 1 + 100 ) ×100 : 2 = 5750\)

\(100 x + 5050 = 5750\)

\( 100 x = 5750 − 5050\)

\(100 x = 700\)

\(x = 700 : 100\)

\(x = 7\)

\(b,\) \(B=\)\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2021^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2020}+2021\)

\( B < 1 -\)\(\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2020}-\dfrac{1}{2021}\)

\(B<1-\)\(\dfrac{1}{2021}\)

\(B<\)\(\dfrac{2020}{2021}\)

\(\dfrac{2020}{2021}< 1\)

\(B<1\)

a) (x+1) +(x+2 ) + ...+(x+100)=5750
= 100x + (1+2+3+...+100) = 5750
=100x + 5050 = 5750
--> 100x = 5750-5050=700
--> x=7

b) Ta thấy: 1/2^2 < 1/2.3
                  1/3^2 < 1/3.4
                        ...
                  1/2021^2 < 1/2021.2022
--> B=1/2^2 + 1/3^2 + 1/4^2 + ...+ 1/2021^2 < 1/2.3 + 1/3.4 + ... +1/2021.2022 (1)
     Ta có: 1/2.3 + 1/3.4 + ... +1/2021.2022
         =1/2 - 1/3 + 1/3 - 1/4 + ... + 1/2021 - 1/2022
         =1/2 - 1/2022 < 1 (2)
Từ (1) và (2) --> B<1 (đpcm)
                                                                      < 
                                                                     

ai lam day du dau tien minh se k cho nha

minh can gap lam

Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)

Ta có : \(\frac{1}{2^2}< \frac{1}{1.2}\)

            \(\frac{1}{3^2}< \frac{1}{2.3}\)

            \(\frac{1}{4^2}< \frac{1}{3.4}\)

             ...

            \(\frac{1}{100^2}< \frac{1}{99.100}\)

\(\Rightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(\Rightarrow A< 1-\frac{1}{100}< 1\)

\(\Rightarrow A< 1\)

Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 1\).

\(P=\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}\)

- Có: \(P>\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{100.101}\)

=> \(P>\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{101}\)

=> \(P>\frac{1}{5}-\frac{1}{101}>\frac{1}{6}\)

=> \(P>\frac{1}{6}\)(1)

- Có: \(P< \frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}\)

=> \(P< \frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+....+\frac{1}{99}-\frac{1}{100}\)

=> \(P< \frac{1}{4}-\frac{1}{100}< 14\)(2)

Từ (1) và (2) 

=> \(\frac{1}{6}< P< 14\)(Nếu đề là 1/6 < P < 1/4 thì thay số 14 bằng 1/4 vẫn đúng nhé)

=> Đpcm